Download Link: https://sites.google.com/view/varunpratapsingh/teaching-engagements UNIT-5 Stress and Strain Analysis Simple stress and strain: Introduction, Normal shear stresses, Stress-strain diagrams for ductile and brittle materials, Elastic constants, One dimensional loading of members of varying cross section, Strain energy, Thermal stresses. Compound stress and strains: Introduction, State of plane stress, Principal stress and strain, Mohr’s circle for stress and strain. Pure Bending of Beams: Introduction, Simple bending theory, Stress in beams of different cross sections. Torsion: Introduction, Torsion of Shafts of circular section, Torque and Twist, Shear stress due to Torque.

1. STRENGTH OF MATERIAL
UNIT 5
PART-1
(SIMPLE STRESS AND STRAIN)
By
Varun Pratap Singh

2. TYPES OF LOAD
There are a number of different ways in which load can be applied to a
member. Typical loading types are:
A) Dead/ Static load- Non fluctuating forces generally caused by gravity
B) Live load- Load due to dynamic effect. Load exerted by a lorry on a
bridge
C) Impact load or shock load- Due to sudden blows
D) Fatigue or fluctuating or alternating loads: Magnitude and sign of the
forces changing with time

3. Types of Loads
Load is defined as the set of external forces acting on a mechanism or
engineering structure which arise from service conditions in which the
components work
In the mechanics of the deformable bodies, the following types of loads are
commonly considered:
• Dead loads—static in nature, such as the self-weight of the roof.
• Live loads—fluctuating in nature, do not remain constant- such as a weight
of a vehicle moving on a bridge.
• Tensile loads.
• Compressive loads.
• Shearing loads
Sign convention followed: Tensile forces are positive and compressive negative

4. Classification of Materials
From an engineering point of view, properties concerned with metals are:
1. Elasticity
2. Plasticity
3. Brittleness
4. Malleability
5. Ductility
Many of these properties are contrasting in nature so that a given metal cannot exhibit simultaneously all these properties.
For example, mild steel exhibits the property of elasticity, copper possesses the property of ductility, wrought iron is
malleable, lead is plastic and cast iron is brittle.
Elastic Material
II undergoes a deformation when subjected to an external loading such that the deformation disappears on the removal of
the loading (rubber).
Plastic Material
It undergoes a continuous deformation during the period of loading and the deformation is permanent. It does not regain its
original dimensions on the removal o the loading (aluminum).
Rigid Material
It does not undergo any deformation when subjected to an external loading (glass and cast iron).
Malleability
Materials ability to be hammered out into thin sheets, such as lead, is called malleability.
Brittle Materials
They exhibit relatively small extensions to fracture such as glass and cast iron. There is little or
no necking at fracture for brittle materials

5. STRESS
• Stress is an internal resistance offered by a unit area of the material, from which a
member is made, to an externally applied load. Alternatively, the force per unit
area or intensity of the forces distributed over a given section is called the stress
on that section. The resistance of material or the internal force acting on a unit area
may act in any direction.
Direct or normal stress G is calculated by using the following formula:

6. Types of Stresses
The stress may be normal stress or a shear stress. Normal stress is the stress which
acts in a direction perpendicular to the area. It is represented by ζ (sigma). The
normal stress is further divided into tensile stress and compressive stress.
Types of stress
There are following type of stresses as displayed in figure here and we will discuss
each type of stress in detail with the help of this post. Basically stresses are
classified in to three types.
1. Direct stress or simple stress
2. Indirect stress
3. Combined stress

7. Tensile Stress:
The stress induced in a body, when subjected to two equal and opposite
pulls as shown in Fig. as a result of which there is an increase in length, is
known as tensile stress. The ratio of increase in length to the original length
is known as tensile strain. The tensile stress acts normal to the area and it
pulls on the area.
Fig. (a) shows a bar subjected to a tensile force P at its ends. Consider a
section x-x, which divides the bar into two parts. The part left to the section
x-x, will be in equilibrium if P = Resisting force (R). This is shown in Fig.
(b). Similarly the part right to the section x-x, will be in equilibrium if P =
Resisting force as shown in Fig. (c). This resisting force per unit area is
known as stress or intensity of stress.
Normal stress
Normal stress is basically defined as the stress acting in a direction
perpendicular to the area. Normal stress will be further divided, as we have
seen above, in two types of stresses i.e. tensile stress and compressive
stress.

8. Compressive Stress:
• The stress induced in a body, when subjected to two equal and opposite pushes as shown in Fig. (a)
as a result of which there is a decrease in length of the body, is known as compressive stress. And
the ratio of decrease in length to the original length is known as compressive strain.
The compressive stress acts normal to the area and it pushes on the area.
Let an axial push P is acting on a body in cross-sectional area A. Due to external push P, let the
original length L of the body decreases by dL

9. Shear Stress
The stress induced in a body, when subjected to two equal and opposite forces which are acting
tangentially across the resisting section as shown in Fig. 1.4 as a result of which the body tends to
shear off across the section, is known as shear stress. The corresponding strain is known as shear
strain. The shear stress is the stress which acts tangential to the area. It is represented by τ.

10. Torsional stress
Torsional stress is one special type of stress in a member where one end of member will
be secured and other end of member will be twisted. Let us see one example, let we have
one shaft and its one end is supported by a ball bearing and other end of shaft is free to
rotate.
Let if bearing is seized, then there will be twisting mechanism because one end of
rotating shaft will be fixed due to seized bearing and other end of shaft will be twisted
and therefore there will be produced stress in shaft and that stress will be termed as
torsional stress.
From here, we can also say that torsional stress will exist in rotating members such as
rotating shaft. Torsional stress will be indicated by symbol

11. Bending stress
Let we have one beam which one end is fixed at A and other end is loaded by force P and hence beam is deflected here as
shown in figure.
Bending stress will be determined with the help of following formula as displayed here in following figure.
We will study and analyze each type of strain in detail in our next post.

12. STRAIN
When a body is subjected to some external force, there is some change of dimension
of the body. The ratio of change of dimension of the body to the original dimension
is known as strain. Strain is dimensionless. Strain may be :
• Tensile strain
• Compressive strain,
• Volumetric strain
• Shear strain
If there is some increase in length of a body due to external force, then the ratio of
increase of length to the original length of the body is known as tensile strain. But if
there is some decrease in length of the body, then the ratio of decrease of the length
of the body to the original length is known as compressive strain. The ratio of
change of volume of the body to the original volume is known as volumetric strain.
The strain produced by shear stress is known as shear strain.

13. TYPES OF STRAIN IN STRENGTH OF MATERIALS
Strain is also classified on the basis of type of deformation
Temporary or elastic strain
Permanent or plastic strain
Strain is also classified as mentioned here
Linear strain
Lateral strain
Volumetric strain
Tensile strain
Let us see here the following figure; we have one bar of length L. There are two equal and opposite pulling type of forces P,
acting axially and trying to pull the bar and this pulling action will result increase in the length of the bar under the action of
tensile loading, but diameter of the bar will be reduced under the action of tensile loading.
Therefore we can define the tensile strain as the strain developed in a member due to the pulling action of two equal and opposite
direction of forces. Ԑt is the symbol which is used to represent the tensile strain in a member.
Tensile strain will be determined with the help of following formula.
Tensile strain (Ԑt) = Increase in length (dL) / Original length (L)
Ԑt = dL / L
Types of strain
There are following types of strain and we will discuss each type of strain in detail with the help of this post.
Strain is classified on the basis of the type of loading
Tensile strain
Compressive strain
Shear strain

14. Compressive strain
Let us see here the following figure; we have one bar of length L. There are two equal and opposite push type of loading P, acting
axially and trying to push the bar and this pushing action will result the decrease in the length of the bar under the action of
compressive loading, but diameter of the bar will be increased under the action of compressive loading. Ԑc is the symbol which is
used to represent the tensile strain in a member.
Compressive strain will be determined with the help of following formula.
Compressive strain (Ԑc) = Decrease in length (dL) / Original length (L)
Ԑc = dL / L
Shear strain
Let us see here the following figure; we have two plates and these two plates are connected with each other with the help of a pin
or a rivet as shown in figure. There are two equal and opposite forces (P) acting tangentially across the resisting section.
One force is acting on top plate towards left direction and second force is acting towards right side as shown in figure and hence
such type of loading will try to shear off the body across the resisting section.
This type of loading action will be termed as shear loading and stress developed in material of the body will be termed as shear
stress and corresponding strain will be termed as shear strain.
Shear strain will be determined as displayed in following figure:

15. Shear strain will be determined as displayed in following figure
Temporary or elastic strain
Temporary strain or elastic strain is such type of strain which will be disappeared on removal of external load or we can say that
body will return to its original shape and size.
Permanent or plastic strain
Permanent strain or plastic strain is such type of strain which will not be disappeared on removal of external load or we can say that
body will not return to its original shape and size.
Linear strain
Let us consider one rectangular bar as shown in figure, its length is L and breadth is B. There are two equal and opposite pulling
type of forces P, acting axially and trying to pull the rectangular bar and this pulling action will result increase in the length of the
rectangular bar under the action of tensile loading, but breadth of the rectangular bar will be reduced under the action of tensile
loading.
Linear strain will be basically defined as the ratio of change in length of the body to the original length of the body. Linear strain
might be tensile or compressive depending on the type of loading i.e. tensile loading or compressive loading. Here in this example,
linear strain will be tensile strain because tensile loading is here.

16. Lateral strain
Consider the same figure as displayed above, Lateral strain will be the strain at perpendicular or right angle to the direction of
applied force.
Therefore we can define here lateral strain as; lateral strain will be basically defined as the ratio of change in breadth of the body
to the original breadth of the body.
Volumetric strain
When an object will be subjected with a system of forces, object will undergo through some changes in its dimensions and
hence, volume of that object will also be changed.
Volumetric strain will be defined as the ratio of change in volume of the object to its original volume. Volumetric strain is also
termed as bulk strain.
Ԑv= Change in volume /original volume
Ԑv= dV/V

17. HOOK’S LAW
For elastic bodies, the ratio of stress to strain is constant and is known as Young's modulus or the
modulus of elasticity and is denoted by E, i.e.,
• Strain has no units as it is a ratio. Thus, E has the same units as stress.
• The materials that maintain this ratio are said to obey Hooke s law which states that within elastic limits,
strain is proportional to the stress producing it. The elastic limit of a material is determined by plotting a
tensile test diagram. Young's modulus is the stress required to cause a unit strain.
• Similarly, for elastic materials, the shear strain is found to be proportional to the applied shear
stress within the elastic limit. Modulus of rigidity or shear modulus denoted by G is the ratio of shear
stress to shear strain, i.e.,
• The ratio between the volumetric (Identical) stress and the volumetric strain is called Bulk
modulus of elasticity and is denoted by K.

18. Lateral strain
The strain at right angles to the direction of applied load is known as lateral strain. Let a rectangular bar of length L,
breadth b and depth d is subjected to an axial tensile load P as shown in Fig. The length of the bar will increase while the
breadth and depth will decrease.
Let
δL = Increase in length,
δb= Decrease in breadth, and
δd = Decrease in depth.
Note:
1) If longitudinal strain is tensile, the lateral strains will be compressive.
2) If longitudinal strain is compressive then lateral strains will be tensile.
3) Hence every longitudinal strain in the direction of load is accompanied by lateral strains of opposite kind in all
directions perpendicular to the load.

19. Longitudinal strain
When a body is subjected to an axial tensile load, there is an increase in the length of the body. But at the same time there
is a decrease in other dimensions of the body at right angles to the line of action of the applied load. Thus the body is
having axial deformation and also deformation at right angles to the line of action of the applied load (i.e., lateral
deformation).
The ratio of axial deformation to the original length of the body is known as longitudinal (or linear) strain. The longitudinal
strain is also defined as the deformation of the body per unit length in the direction of the applied load.
Let L = Length of the body,
P = Tensile force acting on the body,
δL = Increase in the length of the body in the direction of P.
Then,

20. POISON’S RATIO
The ratio of lateral strain to the longitudinal strain is a constant for a given material, when the material
is stressed within the elastic limit. This ratio is called Poisson's ratio and it is generally denoted by μ
or ν or 1/m. Hence mathematically:
Modular ratio is the ratio of Modulus of Elasticity of one material to Modulus of Elasticity of another material.
Modular ratio

21. LINEAR ELASTICITY AND ELASTIC LIMIT
When an external force acts on a body, the body tends to undergo some deformation. If the external force is
removed and the body comes back to its original shape and size (which means the deformation disappears
completely), the body is known as elastic body. This property, by virtue of which certain materials return back to
their original position after the removal of the external force, is called elasticity. The body will regain its
previous shape and size only when the deformation caused by the external force, is within a certain limit. Thus
there is a limiting value of force up to and within which, the deformation completely disappears on the removal
of the force.
The value of stress corresponding to this limiting force is known as the elastic limit of the material. If the
external force is so large that the stress exceeds the elastic limit, the material loses to some extent its property of
elasticity. If now the force is removed, the material will not return to its original shape and size and there will be
a residual deformation in the material.

22. STRESS – STRAIN RELATIONSHIPS
Figure: Stress-strain curve for structural steel
After conducting tension test on steel we can determine the
following items
• Elastic modulus
• Proportional limit
• Yield stress
• Ultimate stress

24. True Stress-Strain Diagram
In plotting stress-strain diagram, we make use of original area of cross section while computing all stress values
and original length while calculating corresponding strains. In this context it is pertinent to define the
following:
Nominal or Conventional or Engineering Stress
The ratio of load over original area of cross section of a component is nominal stress.
True Stress
The ratio of load over instantaneous area of cross section is true stress. Thus, under tensile load, instantaneous
area is less than original area and under compressive load, instantaneous area is more than original area.
Nominal or Engineering Strain
Strain values are calculated at various intervals of gradually increasing load considering original
gauge length of the specimen, such a strain is nominal or engineering strain. Nominal strain is
change in dimension to corresponding original dimension.
True Strain
As the load keeps on increasing, the gauge length will also keep on varying (e.g., gauge length increases under
tensile loading). If actual length is used in calculating the strain, the strain
obtained is true strain.

25. Stress-Strain Diagram for Other Materials

26. Proof Stress
Most of the metals except steel, do not show well-defined yield point and yet undergoes large strains after the
proportional limit is exceeded. An arbitrary yield stress called proof stress for these metals can be found out by offset
method. On the stress-strain diagram of the metal under consideration, a line is drawn parallel to initial linear part of
the curve (Figure) this line is drawn at a standard offset of strain value, such as 0.002 (0.2%). The intersection of the
offset line and the stress-strain curve (point A in the figure) defines the yield point for the metal and hence yield stress.
Proof stress is not an inherent property of the metal. Proof stress is also called offset yield stress .

27. A very basic equation to calculate FoS is to divide the ultimate (or maximum) stress by the typical (or working) stress.
A FoS of 1 means that a structure or component will fail exactly when it reaches the design load, and cannot support
any additional load.
Factor of Safety

28. Margin of safety
• Many government agencies and industries (such as aerospace) require the use of a margin of
safety (MoS or M.S.) to describe the ratio of the strength of the structure to the requirements.
• M.S. as a measure of structural capability: This definition of margin of safety commonly seen in
textbooks describes what additional load beyond the design load a part can withstand before
failing.
• M.S. as a measure of requirement verification: Many agencies and organizations such
as NASAand AIAA define the margin of safety including the design factor, in other words, the
margin of safety is calculated after applying the design factor. In the case of a margin of 0, the part
is at exactly the required strength (the safety factor would equal the design factor).

29. Mechanical Properties of Engineering Materials
• To finalize the material for an engineering product or application, is it important to understand the
mechanical properties of the material. The mechanical properties of a material are those which
affect the mechanical strength and ability of a material to be molded in suitable shape.
• A material's property (or material property) is an intensive property of some material, i.e. a physical
property that does not depend on the amount of the material. Some of the typical mechanical properties
of a material include:
• Strength
• Toughness
• Hardness
• Hardenability
• Brittleness
• Malleability
• Ductility
• Creep and Slip
• Resilience
• Fatigue

30. • Strength
The property of a material which opposes the deformation or breakdown of material in presence of external forces or load. Materials which we
finalize for our engineering products, must have suitable mechanical strength to be capable to work under different mechanical forces or loads.
• Ductility
Ductility is a property of a solid material which indicates that how easily a material gets deformed under tensile stress. Ductility is often
categorized by the ability of material to get stretched into a wire by pulling or drawing. This mechanical property is also an aspect of plasticity
of material and is temperature dependent. With rise in temperature, the ductility of material increases.
• Brittleness
Brittleness of a material indicates that how easily it gets fractured when it is subjected to a force or load. When a brittle material is subjected to
a stress it observes very less energy and gets fractures without significant strain. Brittleness is converse to ductility of material. Brittleness of
material is temperature dependent. Some metals which are ductile at normal temperature become brittle at low temperature.
• Toughness
It is the ability of a material to absorb the energy and gets plastically deformed without fracturing. Its numerical value is determined by the
amount of energy per unit volume. Its unit is Joule/ m3. Value of toughness of a material can be determined by stress-strain characteristics of a
material. For good toughness, materials should have good strength as well as ductility.
For example: brittle materials, having good strength but limited ductility are not tough enough. Conversely, materials having good ductility but
low strength are also not tough enough. Therefore, to be tough, a material should be capable to withstand both high stress and strain.
• Resilience
Resilience is the ability of material to absorb the energy when it is deformed elastically by applying stress and release the energy when stress is
removed. Proof resilience is defined as the maximum energy that can be absorbed without permanent deformation. The modulus of resilience is
defined as the maximum energy that can be absorbed per unit volume without permanent deformation. It can be determined by integrating the
stress-strain cure from zero to elastic limit. Its unit is joule/m3.

31. • Hardness
It is the ability of a material to resist to permanent shape change due to external stress. There are various measure of hardness – Scratch Hardness,
Indentation Hardness and Rebound Hardness.
• Scratch Hardness
Scratch Hardness is the ability of materials to the oppose the scratches to outer surface layer due to external force.
• Indentation Hardness
It is the ability of materials to oppose the dent due to punch of external hard and sharp objects.
• Rebound Hardness
Rebound hardness is also called as dynamic hardness. It is determined by the height of “bounce” of a diamond tipped hammer dropped from a fixed height on the material.
• Hardenability
It is the ability of a material to attain the hardness by heat treatment processing. It is determined by the depth up to which the material becomes
hard. The SI unit of hardenability is meter (similar to length). Hardenability of material is inversely proportional to the weld-ability of material.
• Malleability
Malleability is a property of solid materials which indicates that how easily a material gets deformed under compressive stress. Malleability is
often categorized by the ability of material to be formed in the form of a thin sheet by hammering or rolling. This mechanical property is an aspect
of plasticity of material. Malleability of material is temperature dependent. With rise in temperature, the malleability of material increases.
• Creep and Slip
Creep is the property of a material which indicates the tendency of material to move slowly and deform permanently under the influence of
external mechanical stress. It results due to long time exposure to large external mechanical stress with in limit of yielding. Creep is more severe in
material that are subjected to heat for long time. Slip in material is a plane with high density of atoms.
• Fatigue
Fatigue is the weakening of material caused by the repeated loading of the material. When a material is subjected to cyclic loading, and loading
greater than certain threshold value but much below the strength of material (ultimate tensile strength limit or yield stress limit), microscopic
cracks begin to form at grain boundaries and interfaces. Eventually the crack reaches to a critical size. This crack propagates suddenly and the
structure gets fractured. The shape of structure affects the fatigue very much. Square holes and sharp corners lead to elevated stresses where the
fatigue crack initiates.

32. Mechanical properties (In Short)
• Brittleness: Ability of a material to break or shatter without significant deformation when under stress; opposite of plasticity, examples:
glass, concrete, cast iron, ceramics etc.
• Bulk modulus: Ratio of pressure to volumetric compression (GPa) or ratio of the infinitesimal pressure increase to the resulting relative
decrease of the volume.
• Coefficient of restitution: the ratio of the final to initial relative velocity between two objects after they collide. Range : 0-1, 1 for perfectly
elastic collision.
• Compressive strength: Maximum stress a material can withstand before compressive failure (MPa)
• Creep: The slow and gradual deformation of an object with respect to time
• Ductility: Ability of a material to deform under tensile load (% elongation)
• Durability: Ability to withstand wear, pressure, or damage; hard-wearing.
• Elasticity: Ability of a body to resist a distorting influence or stress and to return to its original size and shape when the stress is removed
• Fatigue limit: Maximum stress a material can withstand under repeated loading (MPa)
• Flexibility: Ability of an object to bend or deform in response to an applied force; pliability; complementary to stiffness
• Flexural modulus
• Flexural strength : The stresses in a material just before it yields.
• Fracture toughness: Ability of a material containing a crack to resist fracture (J/m^2)
• Friction coefficient: The amount of force normal to surface which converts to force resisting relative movement of contacting surfaces
between material pair.

33. Mechanical properties (In Short)
• Hardness: Ability to withstand surface indentation and scratching (e.g. Brinell hardness number)
• Malleability: Ability of the material to be flattened into thin sheets under applications of heavy compressive forces without cracking by hot or cold
working means.
• Mass diffusivity: Ability of one substance to diffuse through another
• Plasticity: Ability of a material to undergo irreversible or permanent deformations without breaking or rupturing; opposite of brittleness
• Poisson's ratio: Ratio of lateral strain to axial strain (no units)
• Resilience: Ability of a material to absorb energy when it is deformed elastically (MPa); combination of strength and elasticity
• Shear modulus: Ratio of shear stress to shear strain (MPa)
• Shear strength: Maximum shear stress a material can withstand
• Slip: A tendency of a material's particles to undergo plastic deformation due to a dislocation motion within the material. Common in Crystals.
• Specific modulus: Modulus per unit volume (MPa/m^3)
• Specific strength: Strength per unit density (Nm/kg)
• Specific weight: Weight per unit volume (N/m^3)
• Stiffness: Ability of an object to resist deformation in response to an applied force; rigidity; complementary to flexibility
• Surface roughness: the deviations in the direction of the normal vector of a real surface from its ideal form.
• Tensile strength: Maximum tensile stress of a material can withstand before failure (MPa)
• Toughness: Ability of a material to absorb energy (or withstand shock) and plastically deform without fracturing (or rupturing); a material's resistance
to fracture when stressed; combination of strength and plasticity
• Viscosity: A fluid's resistance to gradual deformation by tensile or shear stress; thickness
• Yield strength: The stress at which a material starts to yield plastically (MPa)
• Young's modulus: Ratio of linear stress to linear strain (MPa)

34. EXTENSION / SHORTENING OF A BAR
• Consider a prismatic bar of length L and cross-sectional area A subjected to axial force P. We have
the relation
• upon substitution of ε and ζ in that equation, we get
where
E = Young’s Modulus, N/mm2
L = original length , mm
δL = change in length , mm
A = original cross-sectional area, mm2 and
P = axial force , N
The above Eq. can also be written as,

35. PRINCIPLE OF SUPERPOSITION
• When a number of loads are acting on a body, the resulting strain, according to principle of
superposition, will be the algebraic sum of strains caused by individual loads.
• While using this principle for an elastic body which is subjected to a number of direct forces (tensile or
compressive) at different sections along the length of the body, first the free body diagram of individual
section is drawn. Then the deformation of the each section is obtained. The total deformation of the body
will be then equal to the algebraic sum of deformations of the individual sections.

36. BARS WITH CROSS SECTIONS VARYING IN STEPS
• Consider a bar of varying three sections of lengths L1 L2 and L3 having respective areas of cross-
sections A1, A2 and A3 subjected to an axial pull P. Let δL1, δL2, δL3 be the changes in length of
the respective three sections of the bar, then we have
• Now the total elongation of the bar,

37. Strain Energy And Resilience
Strain energy is elastic — that is, the material tends to recover
when the load is removed. ... Resilience is typically expressed as
the modulus of resilience, which is the amount of strain
energy the material can store per unit of volume without causing
permanent deformation.
Strain energy stored in a specimen when strained within elastic
limit is known as Resilience. The maximum energy stored at
elastic limit is known as Proof Resilience. The proof
resilience per unit volume or strain energy per unit volume is
known as Modulus of Resilience.

39. Strain Energy Stored in a Body When The Load is
Gradually Applied
• Let us consider a body which is subjected with tensile load which is increasing gradually up to its
elastic limit from value 0 to value P and therefore deformation or extension of the body is also
increasing from 0 to x and we can see it in following load extension diagram as displayed here.
• σ= Stress developed in the body
• E = Young’s Modulus of elasticity of the material of the body
• A= Cross sectional area of the body
• P = Gradually applied load which is increasing gradually up to its elastic limit from value 0 to value
P
• P = σ. A
• x = Deformation or extension of the body which is also increasing from 0 to x
• L = Length of the body
• V= Volume of the body = L.A
• U = Strain energy stored in the body
• As we have already discussed that when a body will be loaded within its elastic limit, the work done
by the load in deforming the body will be equal to the strain energy stored in the body.

40. Strain energy stored in the body = Work done by the load in deforming the body
Strain energy stored in the body = Area of the load extension curve
Strain energy stored in the body = Area of the triangle ABC
U = (1/2) . AB . BC
U = (1/2) x. P
Let us use the value of P = σ. A, which is determined above
U = (1/2) x. σ. A
Now we will secure the value of extension x in terms of Stress, Length of the body and Young’s modulus of the body by using the
concept of Hook’s Law.
According to Hook’s Law
Within elastic limit, stress applied over an elastic material will be directionally proportional to the strain produced due to external
loading and mathematically we can write above law as mentioned here.
Stress = E. Strain
Where E is Young’s Modulus of elasticity of the material
σ = E. ε
σ = E. (x/L)
x = σ. L/ E
Let use the value of the extension or deformation “x” in strain energy equation and we will have
U = (1/2) (σ. L/ E).σ. A
U = (1/2) (σ2/E) L.A
U = (σ2/2E) V
U = (σ2/2E) V

41. Therefore strain energy stored in a body, when load will be applied gradually, will be given by following equation.
Proof resilience
Proof resilience is basically defined as the maximum strain energy stored in the body. As we know that strain energy stored in the
body will be maximum when body will be loaded up to its elastic limit, therefore if σ could be taken at the elastic limit then we
will have following equation for Proof resilience.
Modulus of resilience
Modulus of resilience is a property of the material which is defined as the proof resilience per unit volume of the body and we can
express as
Modulus of resilience = Proof resilience/Volume of the body

42. STRAIN ENERGY STORED IN A BODY DUE TO
SUDDENLY APPLIED LOAD
• Let us see the load extension diagram as displayed here for this case where body will be subjected with sudden
load and we will find out here the stress induced in the body due to sudden applied load and simultaneously we
will also secure the expression for strain energy for this situation.
Let us go ahead step by step for easy understanding, however if there is any issue we can discuss it in comment box
which is provided below this post.
We have following information from above load extension diagram for body which is subjected with sudden applied
load.
σ = Stress developed in the body due to sudden applied load
E = Young’s Modulus of elasticity of the material of the body
A= Cross sectional area of the body
P = Sudden applied load which will be constant throughout the deformation process of the body
x = Deformation or extension of the body
L = Length of the body
V= Volume of the body = L.A
U = Strain energy stored in the body

43. As we have already discussed that when a body will be loaded within its elastic limit, the work done by the load in deforming the body will be equal to the strain
energy stored in the body.
Strain energy stored in the body = Work done by the load in deforming the body
Strain energy stored in the body = Area of the load extension curve
Strain energy stored in the body = P. x
U = P. x
As we know that maximum strain energy stored in the body U will be provided by the following expression as mentioned here.
Now we will secure the value of extension x in terms of Stress, Length of the body and Young’s modulus of the body by using the concept of Hook’s Law.
According to Hook’s Law
Within elastic limit, stress applied over an elastic material will be directionally proportional to the strain produced due to external loading and mathematically we can
write above law as mentioned here.
Stress = E. Strain
Where E is Young’s Modulus of elasticity of the material
σ = E. ε
σ = E. (x/L)
x = σ. L/ E
Let use the value of the extension or deformation “x” in above equation and we will have
Therefore, we can say here that maximum stress induced in the body due to sudden applied load will be twice the stress induced in the body with same value of load
applied gradually.

44. STRAIN ENERGY STORED DUE TO IMPACT LOADING
• Let us see the following figure, where we can see one vertical bar which is fixed at the upper end and there is collar at the
lower end of the bar. Let us think that one load is being dropped over the collar of the vertical bar from a height of h as
displayed in following figure.
• Such a case of loading could be considered as a body subjected with an impact load and let us consider that there will be
extension in the bar by x due to impact load P.
We will find out here the stress induced in the body due to impact loading and simultaneously we will also discuss for strain
energy for this situation.
Let us go ahead step by step for easy understanding, however if there is any issue we can discuss it in comment box which is
provided below this post.
We have following information from above figure where a body is subjected with an impact load.
•
• σ= Stress developed in the body due to impact load
• E = Young’s Modulus of elasticity of the material of the body
• A= Cross sectional area of the body
• P = Impact load
• x = Deformation or extension of the body i.e. vertical bar
• L = Length of the body i.e. vertical bar
• V= Volume of the body i.e. vertical bar = L.A
• U = Strain energy stored in the body i.e. vertical bar

45. As we have already discussed that when a body will be loaded within its elastic limit, the work done by the load in
deforming the body will be equal to the strain energy stored in the body.
Strain energy stored in the vertical bar = Work done by the load in deforming the vertical bar
Strain energy stored in the vertical bar = Load x Displacement
Strain energy stored in the vertical bar = P. (h + x)
U = P. (h + x)
As we know that strain energy stored in the body U will be provided by the following expression as mentioned here.
Now we will secure the value of extension x in terms of Stress, Length of the body and Young’s modulus of the body by
using the concept of Hook’s Law.

46. According to Hook’s Law
Within elastic limit, stress applied over an elastic material will be directionally proportional to the strain produced due to
external loading and mathematically we can write above law as mentioned here.
Stress = E. Strain
Where E is Young’s Modulus of elasticity of the material
σ = E. ε
σ = E. (x/L)
x = σ. L/ E
Let use the value of the extension or deformation “x” in above equation and we will have.
Once we will have value of the stress (σ) induced in the vertical bar due to impact load, we will easily determine the value of
strain energy stored in the vertical bar due to an impact load.

47. STRAIN ENERGY STORED IN A BODY DUE TO SHEAR STRESS
Let us consider that we have one rectangular block ABCD of length l, height h and breadth b as displayed in following figure. Now
we need to produce here the condition of shear loading of rectangular block in order to develop the shear stress and shear strain.
Therefore, let us consider that bottom face AB of the rectangular block ABCD is fixed and we have applied a shear force P
gradually over the top face CD of the rectangular block as displayed in following figure.
As we can see from above figure that top face CD of the rectangular block ABCD will move by CC1 or DD1 due to the application
of shear force P which is applied gradually over the top face CD of the rectangular block.
We have following information from above diagram for a body which is subjected with shear loading.
P = Shear load applied gradually
τ = Shear stress developed in the body due to shear force P
ϕ = Shear strain = CC1/CB = DD1/DA
G = Modulus of rigidity of the material of the body
l= Length of the body i.e. rectangular block ABCD
b= Breadth of the body i.e. rectangular block ABCD
h= Height of the body i.e. rectangular block ABCD
A= Cross sectional area of the body = b x l
V= Volume of the body = l x b x h
CC1 or DD1 = Shear deformation of the body
U = Strain energy stored in the body i.e. rectangular block ABCD due to shear stress

48. As we have already discussed that when a body will be loaded within its elastic limit, the work done by the load in deforming the
body will be equal to the strain energy stored in the body.
Strain energy stored in the body i.e. rectangular block ABCD = Work done by the load in deforming the body.
Strain energy stored in the body i.e. rectangular block ABCD = Average load x distance
As we have seen above that shear force P is applied gradually over the top face CD of the rectangular block and therefore
average load will be P/2.
Therefore, we will have
Strain energy stored in the body i.e. rectangular block ABCD = P/2 x CC1
Let us recall the concept of shear stress and shear strain and we will find out here that value of shear force P and shear
deformation CC1 and we will use above equation to put the value of shear force P and shear deformation CC1.
Shear stress = Shear load/ Cross sectional area
τ = P/ (b x l)
P = τ x b x l
Shear strain = Shear stress / Modulus of rigidity
ϕ = τ / G
CC1/CB = τ / G
CC1/ h = τ / G
CC1 = τ x h / G

49. Now we have values of shear force P and distance CC1 and we will use above equation of strain energy stored in the body due to
shear stress to put the value of shear force P and shear deformation CC1.
Strain energy stored in the body i.e. rectangular block ABCD = (τ x b x l)/2 x (τ x h / G)
U = τ2 x (b x l x h) /2G
U = τ2 x V /2G

50. Generalized Hooke's Law
The generalized Hooke's Law can be used to predict the deformations caused in a given material by an arbitrary combination
of stresses.
The linear relationship between stress and strain applies for
The generalized Hooke's Law also reveals that strain can exist without stress. For example, if the member is experiencing a
load in the y-direction (which in turn causes a stress in the y-direction), the Hooke's Law shows that strain in the x-direction
does not equal to zero. This is because as material is being pulled outward by the y-plane, the material in the x-plane moves
inward to fill in the space once occupied, just like an elastic band becomes thinner as you try to pull it apart. In this situation,
the x-plane does not have any external force acting on them but they experience a change in length. Therefore, it is valid to
say that strain exist without stress in the x-plane.

51. Volumetric strain will be defined as the ratio of change in volume of the object to its original volume. Volumetric strain is also
termed as bulk strain.
Ԑv= Change in volume /original volume
Ԑv= dV/V
Volumetric strain also known as bulk strain will be determined as following
Ԑv= Change in volume /original volume
Ԑv= dV/V
Ԑv= (y. z. Δx + x. z. Δy + x. y. Δz)/ (xyz)
Ԑv= (Δx/x) + (Δy/y) + (Δz/z)
Ԑv= Ԑx + Ԑy + Ԑz
Let us consider the stresses in various directions and young’s modulus of elasticity
σx= Tensile stress in x-x direction
σy= Tensile stress in y-y direction
σz= Tensile stress in z-z direction
E= Young’s modulus of elasticity
Let us brief here first “Poisson ratio” before going ahead
Poisson ratio = Lateral strain /Linear strain
We can also say that, Lateral strain = Poisson ratio (ν) x Linear strain
As we have already seen that, lateral strain will be opposite in sign to linear strain and therefore above equation will be
written as following
Lateral strain = - Poisson ratio (ν) x Linear strain
Therefore tensile stress acting in x-x direction i.e. σx will produce a tensile strain (σx /E) in x-x direction and lateral
strain (ν. σx /E) in y-y direction and z-z direction.

52. Similarly, tensile stress acting in y-y direction i.e. σy will produce a tensile strain (σy /E) in y-y direction and lateral strain
(ν. σy /E) in x-x direction and z-z direction.
Similarly, tensile stress acting in z-z direction i.e. σz will produce a tensile strain (σz /E) in z-z direction and lateral strain
(ν. σz /E) in x-x direction and y-y direction.
Strain produced in x-x direction
σx will produce a tensile strain (σx /E) in x-x direction
σy will produce a lateral strain i.e. compressive strain here (ν. σy /E) in x-x direction
σz will produce a lateral strain i.e. compressive strain here (ν. σz /E) in x-x direction
Total strain produced in x-x direction
Ԑx = (σx /E) - (ν. σy /E) - (ν. σz /E)
Ԑx = (σx /E) - ν [(σy + σz)/E]
Total strain produced in y-y direction
Ԑy = (σy /E) - ν [(σx + σz)/E]
Total strain produced in z-z direction
Ԑz = (σz /E) - ν [(σx + σy)/E]
Let us add total strain produced in each direction to find out the volumetric strain for a rectangular bar subjected
with three forces mutually perpendicular with each other
Volumetric strain, Ԑv= Ԑx + Ԑy + Ԑz
Volumetric strain, Ԑv= [(σx + σy+ σz)/E-2 ν (σx + σy+ σz)/E]
Volumetric strain, Ԑv= (σx + σy+ σz) (1-2 ν)/E

53. Young’s modulus, bulk modulus and Rigidity modulus of an elastic solid are together called Elastic constants. When a
deforming force is acting on a solid, it results in the change in its original dimension. In such cases, we can use the relation
between elastic constants to understand the magnitude of deformation.
Elastic constant formula
Where,
•K is the Bulk modulus
•G is shear modulus or modulus of rigidity.
•E is Young’s modulus or modulus of Elasticity.
Individually Young’s modulus and bulk modulus and modulus of rigidity are related as-
Derivation of relation between elastic constants
We can derive the elastic constant’s relation by combining the mathematical expressions relating terms individually.
•Young modulus can be expressed using Bulk modulus and Poisson’s ratio as –
E=3K(1−2μ)
•Similarly, Young’s modulus can also be expressed using rigidity modulus and Poisson’s ratio as-E=2G(1+2μ)
•Combining the above two-equation and solving them to eliminate Poisson’s ratio.
Relation Between Elastic Constants
(Young’s modulus, bulk modulus and Rigidity modulus)
Formula SI Units
The relation between modulus of elasticity and modulus of rigidity E=2G(1+μ) N/m2 or pascal(Pa)
The relation between Young’s modulus and bulk modulus E=3K(1−2μ) N/m2 or pascal(Pa)
E=9KG/(3K+G)

54. Derive relationship between the Elastic
constants, i. e. E, K and G.

57. WHAT IS THERMAL STRESS AND STRAIN?
First we will understand here the concept of thermal stress and after that we will see here, in this post, thermal strain.
As we know that if we are going to heat any material, there will be increase in temperature of the material and hence there
will be increase in dimensions of the material. Similarly if we are going to cool the material, there will be decrease in
temperature of the material and hence there will be decrease in dimensions of the material.
Let us consider that we are going to heat or cool any material, as we have discussed, there will be changes in temperature of
the material and therefore there will be changes in dimensions of the material.
We can simply say that there will be free expansion or free contraction in the material according to the rising or lowering of
temperature of the material. If free expansion or free contraction of the material due to change in temperature is restricted
partially or completely, there will be stress induced in the material and this stress will be termed as thermal stress.
We must note it here that if free expansion or free contraction of the material due to change in temperature is not restricted i.e.
expansion or contraction of the material is allowed, there will no stress developed in the material.
So what we have concluded for thermal stress and thermal strain?
Thermal stress will be basically defined as the stress developed in the material due to change in temperature and free
expansion or free contraction of the material, due to change in temperature, will be restricted.

58. Respective strain developed in the material will be termed as the thermal strain.
Derivation for thermal stress and thermal strain formula
Let us consider one metal bar AB as shown in following figure. Let us assume that initial temperature of the
metal bar is T1 and we are heating the metal bar, as shown in figure, at one end to achieve the final temperature
of the metal bar T2. Let us assume that initial length of the metal bar is L.
As we have discussed above that if we are going to heat the material, there will be increase in temperature of the
material and therefore there will be increase in dimensions of the material too.
Increase in temperature of the metal bar, ΔT = T2-T1
Let us think that free expansion of the material, due to rise in temperature, is not restricted i.e. free expansion of
the material is allowed and it is displayed in figure as BC.
Let us assume that free expansion, as a result of increase in temperature of the metal bar due to heating, is δL.
Free expansion will be given by the following formula as displayed here.
δL = α. L. ΔT
Thermal strain will be determined, as written here, with the help of following result
Thermal strain, Ԑ = δL/L
Thermal strain, Ԑ = α. ΔT
Let us think that if one load P will be exerted at B as shown in figure and hence as we have discussed above,
free expansion or free contraction of the material due to change in temperature will be restricted and hence there
will be stress induced in the material i.e. thermal stress will be developed in the material.
Thermal stress, σ = thermal strain x Young’s modulus of elasticity
We must have to note it here that thermal stress is also known as temperature stress and similarly thermal strain
is also termed as temperature strain.

59. WORKED EXAMPLES

73. The following observations were made during a tensile test on a mild steel specimen of 40 mm diameter and 200 mm long:
Elongation with 40,000 N load (within the limit of proportionality) = 0.0304 mm, Yield load = 165,000 N, Maximum load =
245,000 N, Length of the specimen at fracture = 252 mm, Determine the yield stress, the modulus of elasticity, the ultimate
stress and the percentage elongation.
Solution
Given:
Diameter of the specimen = 40 mm
Length of the specimen = 200 mm
Load = 40,000 N
Elongation within the limit of proportionality = 0.0304 mm
Yield load = 165,000 N
Maximum load = 245,000 N
Final length of the specimen = 252 mm
To find the yield stress:
Using the relation for yield stress, we have
To find the modulus of elasticity:
Consider the load within the proportionality Limit.
Then, stress is given by
To find the ultimate stress:
Using the relation for ultimate stress, we have
To find the percentage elongation:
Using the relation, we have Percentage elongation

74. The bar shown in Fig. is subjected to a tensile load of 60 kN. Find the diameter of the middle portion of the bar if the stress is
limited to 120 N/mm2. Also find the length of the middle portion if the total elongation of the bar is 0.12 mm. Take E = 2 x 105
N/mm2.
Solution
To find the diameter at the middle portion of the bar:
Stress in the middle portion of the bar is given by
To find the length of the middle portion of the bar:
Let the length of the middle portion of the bar be x
Stress in the end portion is given by
Also, total elongation = elongation of the end portion + elongation of the middle portion = 0.12
mm

75. Figure shows the bar AB of uniform cross-sectional area is acted upon by several forces. Find the deformation of the bar,
assuming E = 2 x 105 N/mm2.
Solution: The free body diagram (F.B.D.) of individual sections is shown in Figure.

76. A steel bar ABCD of varying cross-section is subjected to the axial forces as shown in Fig. Find the value of P for equilibrium.
If the modulus of elasticity E = 2.1 x 105 N/mm2, determine the elongation of the bar
Solution:
From the equilibrium condition:
Σ Fx=0
+8000 -10,000 + P - 5000 = 0
P = 15,000 - 8000 = 7000 N
To find the elongation of the bar:
Consider the free body diagram (F.B.D.) of the bar,

77. A vertical prismatic bar is fastened at its upper end and supported at the lower end by an unyielding floor as shown in Fig.
Determine the reaction R exerted by the floor of the bar if external loads P1 = 1500 N and P2 = 3000 N are applied at the
intermediate points shown.
Solution
Let A be the cross-sectional area of the bar, and E be the modulus of elasticity.
Elongation of AD = elongation of AB + elongation of BC + elongation of the bar CD
Since there is a rigid support at D, there is a reaction R at
D which causes contraction of AD, i.e.
As there is no change in the length of the bar AD, we have

78. REFERENCES:
1) A Textbook of Strength of Materials By R. K. Bansal
2) Fundamentals Of Strength Of Materials By P. N. Chandramouli
3) Strength of Materials By B K Sarkar
4) Strength of Materials By R. K. Rajput
5) Strength of Materials, S. S. Rattan, Tata McGraw-Hill Education.
Disclaimer:
The course materials used in this document are the contents of the textbooks which are referenced above. I don’t own the
copyright of this document.