Som complete unit 01 notes

simplae stress-strain
thermal stress-strain
poisson's ratio
compound stresses

1. STRENGTH OF MATERIAL
LECTURE-01
CONTENTS:-
•Objective of Subject
•Basics Introduction
•Stress-Strain Diagram of
Ductile and Brittle Materials
PROF. SANJEEV GUPTA

3. BASIC INTRODUCTION
Stress: When a material is subjected to an external force, a resisting force is
set up in the component. The internal resistance force per unit area acting on a
material is called the stress at a point. It is a tensor quantity having unit of N/m
2
or
Pascal.

4. Types of stress:
1. Simple or direct stress (Tension, Compression, Shear)
2. Indirect stress (Bending, Torsion)
3. Combined Stress (Combination of 1 & 2)

5. Strain(e) :When a body is subjected to some external force, there is some change of
dimension of the body. The ratio of change in dimension of the body to the original
dimension is known as strain.
Or, The strain (e) is the deformation produced by stress. Strain is dimensionless.
There are mainly four type of strain
1. Tensile strain
2. Compressive strain
3. Volumetric strain
4. Shear strain
A load may be defined as the combined effect of external forces acting on a body. The load
is applied on the body whereas stress is induced in the material of the body. The loads may
be classified as
1. Tensile load 2. Compressive load
3. Torsional load or Twisting load 4. Bending load
5. Shearing loads

6. Ductile Brittle
Ductile materials can be drawn into wires
by stretching.
Brittle materials break, crack or snap easily.
Ductile materials show deformation. Brittle materials do not show deformation.
Ductility is affected by temperature. Brittleness is affected by pressure (or stress).
Major examples for ductile materials are
metals.
Examples of brittle materials include
ceramic and glass.
Difference between Ductile and Brittle
Stress-Strain Diagram of Ductile and
Brittle Materials

7. Hook's Law
According to Hook‟s law the stress is directly proportional to strain i.e.
Normal Stress (ζ) ∝ Normal Strain (ε)
The constant is known as elastic constant
Normal stress/ Normal strain = Young's modulus or Modulus of elasticity (E)
Shear stress/ Shear strain = Shear modulus or Modulus of Rigidity (G)
Direct stress/ Volumetric strain = Bulk modulus (K)

9. Where K is the strength coefficient
n is the strain hardening exponent
n = 0 perfectly plastic solid
n = 1 elastic solid For most metals, 0.1< n < 0.5

10. From Stress Strain diagram

11. STRENGTH OF MATERIAL
LECTURE-02
CONTENTS:-
•Mechanical Properties of
Materials
•Extension / Shortening of a
bar
•Basics Numerical
•Elongation of bar due to self
weight
PROF. SANJEEV GUPTA

12. Mechanic
al
Properties
Strength
Ductility
Hardness
Creep
Fatigue
Resilience
Toughness
Brittleness
Malleabilit
y
Stiffness
Plasticity
Elasticity

13. Elasticity: Elasticity is the property by virtue of which a material is deformed under the load
and is enabled to return to its original dimension when the load is removed.
Plasticity: Plasticity is the converse of elasticity. A material in the plastic state is
permanently deformed by the application of load and it has no tendency to recover. The
characteristic of the material by which it undergoes inelastic strains beyond those at the
elastic limit is known as plasticity.
Ductility: Ductility is the characteristic which permits a material to be drawn out
longitudinally to a reduced section, under the action of a tensile force (large deformation).
Brittleness: Brittleness implies the lack of ductility. A material is said to be brittle when it
cannot be drawn out by tension to the smaller section.
Properties of Materials
Malleability: Malleability is a property of a material which permits the material to be
extended in all directions without rapture. A malleable material possesses a high degree of
plasticity, but not necessarily great strength.

14. Toughness: Toughness is the property of a material which enables it to absorb energy without
fracture.
Note- Toughness = the ability to absorb energy up to fracture = the total area under the
strain-stress curve up to fracture. Can be measured by an impact test (Units: the energy per
unit volume, e.g. J/m
3
)
Hardness: Hardness is the ability of a material to resist indentation or surface abrasion. The
Brinell hardness test is used to check the hardness.
Creep is defined as time dependent deformation when material is under constant loading,
generally it is occur due to variation in grain structure of the material
Fatigue-Fatigue is a situation in which component is subjected to cyclic loading

15. Resilience- It is the property of a material to absorb energy when it is deformed elastically
and then, upon unloading, to have this energy recovered. In other words, it is the maximum
energy per volume that can be elastically stored.
Strength: The st
r
ength of a material enables it to resist fracture under load.
Stiffness is the resistance of an elastic body to deflection or deformation by an applied
force - and can be expressed as
k = F / δ
where
k = stiffness (N/m)
F = applied force (N)
δ = extension, deflection (m)

16. EXTENSION / SHORTENING OF A BAR
Consider a prismatic bar of length L and cross-sectional area A subjected to axial force F. 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,

17. Que.1 A circular rod of diameter 20 m and 500 m long is subjected to tensile force of
45kN. The modulus of elasticity for steel may be taken as 200 kN/m
2
. Find stress, strain
and elongation of bar due to applied load.
Solution- Given data:
D = 20m
L = 500m
P = 45KN = 45000N
E = 200 KN/m
2
= (200 × 1000 N/mm
2
= 200000 N/m
2

18. Que-02 The following observations were made during a tensile test on a mild steel
specimen 40 mm in diameter and 200 mm long. Elongation with 40 kN load (within
limit of proportionality),
δL = 0.0304 mm
Yield load =161 KN
Maximum load = 242 KN
Length of specimen at fracture = 249 mm
Determine:
(i) Young's modulus of elasticity
(ii) Yield point stress
(iii) Ultimate stress
(iv) Percentage elongation.

20. ELONGATION OF BAR DUE TO SELF WEIGHT
Consider a prismatic or circular bar of cross-
sectional area A and length L hanging freely
under its own weight as shown in Fig. This
circular bar experiences zero load at the free
end and maximum load at the top. Weight of a
body is given by the product of density and
volume. Let γ be the density of the material.
Consider a small section of thickness dx at a
distance x from the free end.
Where Wx = weight of the portion below the section = ϒ A x
The extension of the entire bar can be obtained by integrating above Eq.

21. If W is the total weight of the bar, then
Note:
The deformation of the bar under its own
weight is equal to the half of the deformation, if
the body is subjected to the direct load equal to
the weight of the body.

22. STRENGTH OF MATERIAL
LECTURE-03
CONTENTS:-
•Extension of a Tapered bar
•Principle of Superposition
•Bars are Series and Parallel
•Compound bars
PROF. SANJEEV GUPTA

23. EXTENSION OF A TAPERED BAR
Consider a bar of length L tapering from d
2 to d
1 and subjected to axial tensile load P. as
shown in Fig. At a distance x from diameter d
2 , the diameter of the bar is

25. 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.

26. Case-1 Consider a bar of different length, area of cross-section and diameter, subjected to
a axial load of „P‟ as shown in figure-1.
P
= Bar is subjected here with an axial Load
A
1, A
2 and A
3 = Area of cross section of section 1, section 2 and section 3 respectively
L
1, L
2 and L
3 = Length of section 1, section 2 and section 3 respectively
ζ
1, ζ
2 and ζ
3 = Stress induced for the section 1, section 2 and section 3 respectively
ε
1, ε
2 and ε
3 = Strain developed for the section 1, section 2 and section 3 respectively
E= Young‟s Modulus of the bar
Total Change in length, dL = Sum of Change in length of Section 1, Section 2 and Section 3.
dL = dL
1 + dL
2 +dL
3
BARS IN SERIES AND PARALLEL

27. Stress and strain produced for the section 1
Stress, ζ
1= P / A
1
Strain, ε
1 = ζ
1/E
Strain, ε
1 = P / A
1E
Similarly stress and strain produced for the section 2
Stress, ζ
2= P / A
2
Strain, ε
2 = ζ
2/E
Strain, ε
2 = P / A
2E
Similarly stress and strain produced for the section 3
Stress, ζ
3= P / A
3
Strain, ε
3 = ζ
3/E
Strain, ε
3 = P / A
3E
Stepped bar

28. Now we will determine change in length for each section with the help of definition
of strain
ΔL
1= ε
1L
1
ΔL
2= ε
2L
2
ΔL
3= ε
3L
3
Total change in length of the bar
Let us consider that Young‟s modulus of elasticity of each section is different, in this
situation we will have following equation to determine the total change in length of the bar.

29. 1 2
P
Case-II Bar made up of 2 or more bars of equal length but of different material rigidly
fixed with each other
COMPOSITE BAR

30. Conditions Should be Remembers For Series and Parallel bars
(a) For Series bars

31. (b) For Parallel bars

32. Que.3 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/mm
2
. Also find the length of
the middle portion if the total elongation of the bar is 0.12 mm. Take E = 2 x 10
5
N/mm
2
.
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

33. Also, total elongation = elongation of the end portion + elongation of the middle portion
= 0.12 mm

34. STRENGTH OF MATERIAL
LECTURE-04
CONTENTS:-
•Numerical on Series and
Parallel bars
PROF. SANJEEV GUPTA

35. IMPORTANT FORMULAS

37. Que.4 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 10
5
N/mm
2
.
Solution: The free body diagram (F.B.D.) of individual sections is shown in Figure.
Equilibrium condition
Load towards left = Load towards right
10+5 = 3+12 = 15

39. In the case of Parallel bars
P = P
1 + P
2 + P
3 …. …(1)
ΔL
1 = ΔL
2 …… (2)

41. Que 6 A solid steel bar 0.5 m long and 50 mm diameter is placed inside an aluminium tube having 60
mm inside and 80 mm outside diameter. The aluminium tube is 0.2 mm longer than the steel bar. An
axial load of 500 kN is applied to the bar and tube through rigid cover plates as shown in Fig. Find
the stresses developed in steel bar and aluminium tube. E5 = 220 GPa, Ea = 70 GPa.

43. STRENGTH OF MATERIAL
LECTURE-05
CONTENTS:-
•Temperature Stress and
Strain
•Compound Section
•Numerical based on
Temperature stress and
strain
PROF. SANJEEV GUPTA

44. TEMPERATURE STRESS & STRAIN
L= length of a bar of uniform cross-section
t
1 = Initial temperature of the bar
t
2 = final temperature of the bar and
α = Co-efficient of linear expansion
The extension of the bar due to rise in temperature will be

= α(t
2- t
1).l
= αΔt.l
As the material is heated it expand and as cooled it contract.

45. Compound Section

50. STRENGTH OF MATERIAL
LECTURE-06
CONTENTS:-
•Strain Analysis-Poisson’s
ratio
•Two and Three dimensions
Stress system
•Volumetric Strain
PROF. SANJEEV GUPTA

54. VOLUMETRIC STRAIN
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.
Ԑ
v= Change in volume /original volume
Ԑ
v= dV/V
We can also say that we have following initial
dimensions of the rectangular bar
x = Length of the rectangular bar
y = Width of the rectangular bar
z = Thickness or depth of the rectangular bar
Volume of the rectangular bar, V = xyz
Δx=Change in length of the rectangular bar
Δy=Change in width of the rectangular bar
Δz= Change in thickness or depth of the rectangular bar
ΔV= Change in volume of the rectangular bar

55. Let us determine the final dimensions of the rectangular bar
Final volume of the rectangular bar = (x+ Δx). (y + Δy). (z + Δz)
Let us ignore the product of small quantities and we will have
Final volume of the rectangular bar = xyz + y. z. Δx + x. z. Δy + x. y. Δz
Let us determine the change in volume of the rectangular bar
Change in volume of the rectangular bar = Final volume – initial volume
ΔV= (xyz + y. z. Δx + x. z. Δy + x. y. Δz) - xyz
ΔV= y. z. Δx + x. z. Δy + x. y. Δz
Volumetric strain also known as bulk strain will be determined as following
Ԑ
v= Change in volume /original volume
Ԑ
v= dV/V = (y. z. Δx + x. z. Δy + x. y. Δz)/ (xyz)
Ԑ
v= (Δx/x) + (Δy/y) + (Δz/z)
Ԑ
v= Ԑ
x + Ԑ
y + Ԑ
z

56. Let us consider the stresses in various directions and young‟s modulus of elasticity
ζ
x= Tensile stress in x-x direction
ζ
y= Tensile stress in y-y direction
ζ
z= Tensile stress in z-z direction
E= Young‟s modulus of elasticity
Poisson ratio = Lateral strain /Linear strain
Total strain produced in x-x direction
Ԑ
x = (ζ
x /E) - (µ. ζ
y /E) - (. µζ
z /E)
Ԑ
x = (ζ
x /E) - µ [(ζ
y + ζ
z)/E]
Total strain produced in y-y direction
Ԑ
y = (ζ
y /E) - µ [(ζ
x + ζ
z)/E]
Total strain produced in z-z direction
Ԑ
z = (ζ
z /E) - µ [(ζ
x + ζ
y)/E]

57. Volumetric strain, Ԑ
v= Ԑ
x + Ԑ
y + Ԑ
z
Volumetric strain, Ԑ
v= [(ζ
x + ζ
y+ ζ
z)/E-2 µ (ζ
x + ζ
y+ ζ
z)/E]
Volumetric strain, Ԑ
v= (σ
x + σ
y+ σ
z) (1-2 µ)/E

58. STRENGTH OF MATERIAL
LECTURE-07
CONTENTS:-
•Bulk Modulus and Shear
Modulus
•Relationship between Elastic
Constant.
PROF. SANJEEV GUPTA

59. The relative change in the volume of a body produced by a unit compressive or tensile
stress acting uniformly over its surface.
OR
When a body is subjected to three mutually perpendicular like stresses of equal intensity
(ζ), the ratio of direct stress (σ) to the corresponding volumetric strain (εv ) is defined as
the bulk modulus K for the material of the body.

Bulk Modulus

60. Shear Modulus or Modulus of Rigidity
The shear modules or modulus of rigidity (also called the modulus of transverse
elasticity) expresses the relation between shear stress and shear strain. It has been found
experimentally that, within elastic limit, shear stress (τ) is proportional to the shear strain
Where G = modulus of rigidity
(Also sometimes denoted by symbol N or C)

61. There are Three Types of Modulus
1.Young Modulus or Modulus of Elasticity(E)
E = ζ/ ε
2. Bulk Modulus (K)

3. Shear Modulus or Modulus of Rigidity (C/N/G)

62. Relationship between Elastic Constant
(a) Relationship between young modulus(E) and shear modulus (C/N/G)
LMST is a solid cube subjected to a shearing
force F. Let η be the shear stress produced in the
faces MS and LT due to this shearing force. The
complementary shear stress consequently
produced in the faces ML and ST is also η. Due
to the shearing load, the cube is distorted to
LM‟S‟T, and as such the edge M moves to M‟, S
to S‟ and diagonal LS to L‟S.

64. ...(A)

65. ...(B)

The relation between E,K and C can be established by eliminating m from the
equation (A) and (B)

66. From equation (A)

67. E = Young‟s Modulus, also known as Modulus of Elasticity
G = Shear Modulus, also known as Modulus of Rigidity
K = Bulk Modulus
µ = Poisson‟s Ratio = (1/m)

Shear Modulus from Young’s Modulus
Shear Modulus from the Bulk Modulus
Bulk Modulus from Young’s Modulus
Bulk Modulus from the Shear Modulus
Young’s Modulus from the Shear Modulus
Young’s Modulus from the Bulk Modulus

68. STRENGTH OF MATERIAL
LECTURE-08
CONTENTS:-
•Numerical based on
Poisson’s ratio and Elastic
constant.
PROF. SANJEEV GUPTA

69. IMPORTANT FORMULA
Volumetric strain, Ԑ
v= (σ
x + σ
y+ σ
z) (1-2 µ)/E
Poisson’s Ratio =

72. Question- A metallic bar 300 mm x 100 mm x 40 mm is subjected to a force of 5 kN
(tensile), 6 kN (tensile) and 4 kN (tensile) along x, y and z directions respectively.
Determine the change in the volume of the block. Take E = 2 x 10
5
N/mm
2
and Poisson's
ratio = 0.25.
Solution:
Given: Dimensions of bar
= 300 mm x 100 mm x 40 mm
x = 300 mm, y = 100 mm and z = 40 mm
V= xyz = 300 x 100x40 = 1200000 mm
3

74. STRENGTH OF MATERIAL
LECTURE-09
CONTENTS:- COMBINED STRESSES
•Principle planes, Principle
stress
•Derivation of normal and
shear stress on inclined plane
PROF. SANJEEV GUPTA
Types of stress:
1. Simple or direct stress (Tension,
Compression, Shear)
2. Indirect stress (Bending, Torsion)
3. Combined Stress (Combination of 1 & 2)

75. Principle stress, Principle planes
The plane which have no shear stress are known as Principal Plane. These planes
carry only normal stress.
The normal stresses acting on a principal plane are known as Principal Stress.

To determine Stresses on a oblique planes there are two methods
1. Analytical Method 2. Graphical Method (Mohr‟s Circle)

The following Four types of stressed conditions in an element are considered.
a) Uniaxial direct stresses
b) Bi-axial direct stresses
c) Pure shear stress
c) Bi-Axial and Shear stresses condition

76. θ ζ
x ζ
x
A B
C

θ ζ
x ζ
x
A B
C
ζ
y
ζ
y
ηxy

θ ζ
x ζ
x
A B
C
ζ
y
ζ
y
ηxy
ηxy
ηxy
ηxy
ηxy
ηxy

θ
A B
C
ηxy
(A) Uniaxial direct stresses (B) Biaxial direct stresses
(C) Pure shear stress (D) Bi-Axial and Shear stresses condition

77. Uniaxial direct stresses
θ

81. A
B
C
θ
θ
ζ
n
η
θ
ζ
x
ηxy
ζ
y
ηxy
ηxy

θ ζ
x ζ
x
A B
C
ζ
y
ζ
y
ηxy
ηxy
ηxy
(A)
(B)
(C)
(D)
(E)
(D) Bi-Axial and Shear stresses condition
θ
θ ζ
x
ηxy. BC
ζ
x
BC Cosθ

ηxy BC Sinθ

ζ
x
BC Sinθ

ηxy BC Cosθ

θ
ζ
y
AB
θ
ηxy AB Cosθ

ηxy AB

ζ
y
AB Cosθ
ζ
y
AB Sinθ
ηxy AB Sinθ

Sinθ = AB/AC and Cosθ = BC/AC

82. (C)
(D)
Resolving the forces perpendicular to the plane AC
Resolving the forces along the plane AC
θ
θ ζ
x
ηxy. BC
ζ
x
BC Cosθ

ηxy BC Sinθ

ζ
x
BC Sinθ

ηxy BC Cosθ

Sinθ = AB/AC and Cosθ = BC/AC
θ
ζ
y
AB
θ
ηxy AB Cosθ

ηxy AB

ζ
y
AB Cosθ
ζ
y
AB Sinθ
ηxy AB Sinθ

83. To determine the plane having max and min values of direct stress differentiate equation
(1) w.r.t θ and equal to zero

91. Stress Condition

Normal Stress (ζ
θ )

Tangential Stress (η
θ)
Uniaxial direct stresses

Cos 2θ = 2cos
2
θ-1

Bi-axial direct stresses

Pure shear stress


θ ζ
x ζ
x
A B
C

θ ζ
x ζ
x
A B
C
ζ
y
ζ
y
ηx
y
ηxy
ηxy

θ
A B
C
ηxy

92. Angle between the resultant stresses „ζ
r' and the given plane is θ (angle of obliquity)
then
Resultant Stress ζ
r

93. STRENGTH OF MATERIAL
LECTURE-10
CONTENTS:-
•Graphical Method-Mohr’s
circle
•Construction of Mohr’s
Circle- All conditions


PROF. SANJEEV GUPTA

94. Graphical Methods
1. Mohr‟s circle construction for “Like Stresses”
2. Mohr‟s circle construction for “Unlike Stresses”
3. Mohr‟s circle construction for “Two Perpendicular direct stresses with state of simple
Shear stress”
4. Mohr‟s circle construction for “Principal Stresses”
Mohr’s circle

95. A
ζ
x
B ζ
y C
D
E
2θ ϕ
ζ
r
ζ
n
O X
Y
η
θ
1. Mohr’s circle construction for “Like Stresses
FROM STRESS DIAGRAM

96. 2. Mohr’s circle construction for “Unlike Stresses”

97. 3. Mohr’s circle construction for “Two Perpendicular direct stresses with state of
simple Shear stress”
O X
Y
A
ζ
x
B ζ
y
S
T
C
C
B
A
2θ
D
E ϕ
ζ
r
ζ
n
η
θ
η
max
L M β

101. Mohr’s circle construction for “Principal Stresses”

102. STRENGTH OF MATERIAL
LECTURE-11
CONTENTS:-
•Numericals based on Mohr’s
circle
PROF. SANJEEV GUPTA

103. Que. The principal stresses at appoint across two perpendicular planes are 75 MN/m
2
(Tensile) and 35 MN/m
2
(Tensile). Find the normal, tangential stresses and resultant stress
and its obliquity on a plane at 20
0
with the major principal plane.

θ ζ
x
ζ
x
A B
C
ζ
y
ζ
y
= 35 MN/m
2

= 35 MN/m
2

= 70 MN/m
2

= 70 MN/m
2

= 20
0

Solution (ANALYTICAL METHOD)
(1)Normal Stress
(2)Tangential Stress
(3)Resultant Stress
(4)Angle of Obliquity
Given
=75 MN/m
2
(Tensile) and
= 35 MN/m
2
(Tensile)
ζ
x
ζ
y
Θ = 20
0

(1)ζ
n
(2) η
(3)ζ
r
(4)

104. Solution (GRAPHICAL METHOD)
Take a suitable scale 10 MN/m
2
= 1cm
C
D
E A B O

105. = 50 MN/m
2

= 50 MN/m
2

= 30 MN/m
2

= 30 MN/m
2

= 30 MN/m
2

= 30 MN/m
2

= 30 MN/m
2

= 30 MN/m
2

Que. The state of stress at a point of a machine is shown in fig. Determine the principal
stresses and maximum shear stress and its inclination..
Find
(1)Principal stress ζ
1 ,
ζ
2
(2) Maximum shear stress η
max
(3) Inclination θ
Solution
Given
=50 MN/m
2
(Tensile) and
= 30 MN/m
2
(Compressive)
ζ
x
ζ
y
η
xy = 30 MN/m
2

107. O
S
C
A B C O
T
S
β=2θ
β
T
η
max
ζ
x
ζ
y
M
L
ζ
1 ζ
2