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Branches of Mechanics
Engineering Mechanics
Mechanics of Solids
Rigid Bodies
Statics Dynamics
Mechanics of Fluids
Deformable
Bodies
Strength of
Materials
Ideal Viscous Compressible
Fluids
Fluids Fluids
Theory of
Elasticity
Theory of
Plasticity
1
2
COURSE CONTENT IN BRIEF
1. Simple stress and strain
2. Stresses due to bending
3. Stresses due to shearing
4. Stresses due to Torsion in circular shaft
CHAPTER – I
Simple stress & strain
•
•
•
•
•
•
•
•
Normal stress and strain
Hooke’s law
Modulus of elasticity
Tension test on ductile and brittle materials
Factor of safety and allowable stress
Poisson's ratio
Shear stress and shear strain
Modulus of rigidity
4
CHAPTER – I
5
Introduction
The subject strength of materials deals with the relations between
externally applied loads and their internal effects on bodies. The
bodies are no longer assumed to be rigid and the deformations,
however small, are of major interest
The subject, strength of materials or mechanics of materials involves
analytical methods for determining the strength , stiffness
(deformation characteristics), and stability of various load carrying
members.
Alternatively the subject may be called the mechanics of solids.
7
GENERAL CONCEPTS
STRESS
No engineering material is perfectly rigid and hence, when a
material is subjected to external load, it undergoes
deformation.
While undergoing deformation, the particles of the material
offer a resisting force (internal force). When this resisting
force equals applied load the equilibrium condition exists
and hence the deformation stops.
These internal forces maintain the externally applied forces
in equilibrium.
8
STRESS
The internal force resisting the deformation per unit area is
called as stress or intensity of stress.
Stress = internal resisting force / resisting cross sectional area
R

A
9
STRESS
SI unit for stress
N/m2 also designated as a pascal (Pa)
Pa = N/m2
kilopascal, 1kPa = 1000 N/m2
megapascal, 1 MPa = 1×106 N/m2
= 1×106 N/(106mm2) = 1N/mm2
1 MPa = 1 N/mm2
gigapascal, 1GPa = 1×109 N/m2
= 1×103 MPa
= 1×103 N/mm2
10
P
Consider a uniform bar of cross
sectional area A, subjected to a tensile
force P.
P
R
Consider a section AB normal to the
direction of force P
Let R is the total resisting force acting
on the cross section AB.
B
A
STRESS
Then for equilibrium condition,
R
P
R=P
Then from the definition of stress,
normal stress = σ = R/A = P/A
P
Symbol:
σ = Normal Stress
11
STRESS
Direct or Normal Stress:
Intensity of resisting force perpendicular to or normal to the
section is called the normal stress.
Normal stress may be tensile or compressive
Tensile stress:
stresses that cause pulling on the surface of the
section, (particles of the materials tend to pull
apart causing extension in the direction of force)
Compressive stress: stresses that cause pushing on the surface of
the section, (particles of the materials tend to
push together causing shortening in the
direction of force)
12
STRESS
• The resultant of the internal forces for an
axially loaded member is normal to a
section cut perpendicular to the member
axis.
• The force intensity on that section is
defined as the normal stress.
F
  lim
A0 A
P
 ave 
A
13
Example 1
Illustrative Problem
A composite bar consists of an aluminum section rigidly fastened
between a bronze section and a steel section as shown in figure.
Axial loads are applied at the positions indicated. Determine the
stress in each section.
Bronze
A= 120 mm2
Aluminum
A= 180 mm2
13kN
4kN
300mm
400mm
Steel
A= 160 mm2
2kN
7kN
500mm
14
Illustrative Problem
To calculate the stresses, first determine the forces in each
section.
To find the Force in bronze section,
consider a section bb1 as shown in the figure
b
4kN
13kN
2kN
7kN
Bronze
b1
For equilibrium condition algebraic sum of forces on LHS
of the section must be equal to that of RHS
15
Illustrative Problem
b
4kN
13kN
2kN
7kN
Bronze
b1
4kN
Bronze
4kN
(=
13kN
2kN
7kN
)
Force acting on Bronze section is 4kN, tensile
Stress in Bronze section =
Force in Bronze section
Resisting c/s area of the Bronze section
4kN
4 1000 N
2


33
.
33
N
/
mm
=
= 33.33MPa
2
2
120mm
120mm
Tensile stress
16
Force in Aluminum section
4kN
Illustrative Problem
13kN
2kN
7kN
Aluminum
4kN
13kN
9kN
(=
2kN
7kN
Aluminum
Force acting on Aluminum section is 9kN, Compressive
)
17
Force in steel section
4kN
13kN
Illustrative Problem
2kN
7kN
steel
7kN
4kN
13kN
2kN
steel
Force acting on Steel section is 7kN,
Compressive
7kN
18
Illustrative Problem
Stress in Aluminum
section
=
=
Stress in Steel section =
Force in Al section
Resisting cross sectional area of the Al section
9kN
9 1000 N
2


50
N
/
mm
= 50MPa
2
2
180mm
180mm
Compressive stress
Force in Steel section
Resisting cross sectional area of the Steel section
7kN
7 1000 N
2


43
.
75
N
/
mm
= 43.75MPa
= 160mm2 160mm2
Compressive stress
19
STRAIN
STRAIN :
When a load acts on the material it will undergo deformation.
Strain is a measure of deformation produced by the application of
external forces.
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 change in
length by an amount δL, the linear strain produced is defined as,
Linear strain,

L
L
Change in length
=
Original length
Strain is a dimension less quantity.
20
Linear Strain
P
A
2 


2L L

P
   st ress
A
2P P


2A A



L
 normalst rain

L
21
STRESS-STRAIN DIAGRAM
In order to compare the strength of various materials it is necessary
to carry out some standard form of test to establish their relative
properties.
One such test is the standard tensile test in which a circular bar of
uniform cross section is subjected to a gradually increasing tensile
Measurement of change in length over a selected gauge length of the
extensometers.
A graph of load verses extension or stress against strain is drawn as
shown in figure.
22
STRESS-STRAIN DIAGRAM
Proportionality limit
Typical tensile test curve for mild steel
23
STRESS-STRAIN DIAGRAM
Typical tensile test curve for mild steel showing upper yield point and
lower yield point and also the elastic range and plastic range
24
Stress-strain Diagram
Limit of Proportionality :
From the origin O to a point called proportionality limit the stress
strain diagram is a straight line. That is stress is proportional to
strain. Hence proportional limit is the maximum stress up to which
the stress – strain relationship is a straight line and material behaves
elastically.
From this we deduce the well known relation, first postulated by
Robert Hooke in 1678, that stress is proportional to strain.
Beyond this point, the stress is no longer proportional to strain
PP
P 
=
A
Original cross sectional area
25
Stress-strain Diagram
Elastic limit:
It is the stress beyond which the material will not return to its
original shape when unloaded but will retain a permanent
deformation called permanent set. For most practical purposes it can
often be assumed that points corresponding proportional limit and
elastic limit coincide.
Beyond the elastic limit plastic deformation occurs and strains are
not totally recoverable. There will be thus some permanent
PE
E 
=
A
Original cross sectional area
26
Stress-strain Diagram
Yield point:
It is the point at which there is an appreciable elongation or yielding
of the material without any corresponding increase of load.
PY
Y 
=
A
Original cross sectional area
27
Stress-strain Diagram
Ultimate strength:
It is the stress corresponding to maximum load recorded during the
test. It is stress corresponding to maximum ordinate in the stressstrain graph.
PU
U 
=
A
Maximum load taken by the material
Original cross sectional area
28
Stress-strain Diagram
Rupture strength (Nominal Breaking stress):
It is the stress at failure. For most ductile material including
structural steel breaking stress is somewhat lower than ultimate
strength because the rupture strength is computed by dividing the
PB
B 
=
A
Original cross sectional area
True breaking stress,
= Actual cross sectional area
29
Stress-strain Diagram
After yield point the graph becomes much more shallow and covers
a much greater portion of the strain axis than the elastic range.
The capacity of a material to allow these large plastic deformations is
a measure of ductility of the material.
Ductile Materials:
The capacity of a material to allow large extension i.e. the ability to
be drawn out plastically is termed as its ductility. Material with high
ductility are termed ductile material.
Example: Low carbon steel, mild steel, gold, silver, aluminum.
30
Stress-strain Diagram
A measure of ductility is obtained by measurements of the
percentage elongation or percentage reduction in area, defined as,
Percentage elongation =
increase in gauge length (up to fracture)
×100
original gauge length
Reduction in cross sectional area
of necked portion (at fracture)
Percentage reduction in
=
area
original area
×100
Brittle Materials :
A brittle material is one which exhibits relatively small extensions
before fracture so that plastic region of the tensile test graph is
much reduced.
Example: steel with higher carbon content, cast iron, concrete, brick
31
STRESS-STRAIN : DUCTILE MATERIAL
32
Stress-Strain Diagram: Ductile Materials
pounds (lb )
squareinch (in 2 )
1000  lb
ksi 
in 2
psi 
Stress-Strain Diagram: Brittle Materials
Stress-strain diagram for a typical brittle material
33
34
Stress-Strain Test
Machine used to test tensile test specimen
L = gauge length
HOOKE”S LAW
For all practical purposes, up to certain limit the relationship
between normal stress and linear strain may be said to be linear for
all materials
stress (σ) α strain (ε)
stress (σ)
constant
=
strain (ε)
Thomas Young in 1807 introduced a constant of proportionality that
came to be known as Young’s modulus.
stress (σ)
E
=
strain (ε)
=
Young’s Modulus
or
Modulus of Elasticity
35
HOOKE”S LAW
36
Young’s Modulus is defined as the ratio of normal stress to linear
strain within the proportionality limit.
stress (σ)
E = strain (ε) =
P L PL


A L AL
The value of the Young’s modulus is a definite property of a
material
From the experiments, it is known that strain is always a very small
quantity, hence E must be large.
For Mild steel, E = 200GPa = 2×105MPa = 2×105N/mm2
37
Material
Density Young's Modulus
(kg/m3)
109 N/m2
Ultimate Strength Su
106 N/m2
Yield Strength Sy
106 N/m2
Steel
7860
200
400
250
Aluminum
2710
70
110
95
Glass
2190
65
50
...
Concrete
2320
30
40
...
Wood
525
13
50
...
Bone
1900
9
170
...
Polystyrene
1050
3
48
...
Hooke’s Law: Modulus of Elasticity
38
Below the yield stress
  E
E  Youngs Modulus or
Modulus of Elasticity
Strength is affected by alloying, heat
treating, and manufacturing process but
stiffness (Modulus of Elasticity) is not.
Stress-strain diagram for Iron and
Stress-strain diagram
Hard drawn wire materials
Various types of nylon and
polycarbonate
39
Elastic vs. Plastic Behavior
40
• If the strain disappears
when the stress is removed,
the material is said to
behave elastically.
• The largest stress for
which this occurs is called
the elastic limit.
• When the strain does not return to zero after the stress is removed,
the material is said to behave plastically.
Elastic vs. Plastic Behavior
41
For certain materials, for example, high carbon steel and non-ferrous
metals, it is not possible to detect any difference between the upper
and lower yield points and in some cases no yield point exists at all.
In such cases a proof stress is used to indicate beginning of plastic
strain.
Proof stress is the stress corresponding to a fixed permanent strain in
stress-strain diagram.
For example: 0.1% proof stress indicates that stress which, when
removed, produces a permanent strain or “set” of 0.1% of the original
gauge length.
Proof stress
Determination of 0.1%
proof stress
Permanent deformation or “set”
after straining beyond yield point
42
• From Hooke’s Law:
  E


E

P
AE
• From the definition of strain:


L
• Equating and solving for the
deformation,
PL

AE
Pi Li
 
i Ai Ei
43
Example 2
Illustrative Problem
44
A specimen of steel 20mm diameter with a gauge length of
200mm was tested to failure. It undergoes an extension of
0.20mm under a load of 60kN. Load at elastic limit is 120kN.
extension is 50mm and the diameter at fracture is 16mm. Find:
a) Stress at elastic limit
b) Young’s modulus
c) % elongation
d) % reduction in area
e) Ultimate strength
f) Nominal breaking stress
g) True breaking stress
45
Illustrative Problem
Example 2
Solution:
a) Stress at elastic limit,
σE =
Original c/s area
PE
120kN


 381.97 N
2  381.97 MPa
2
mm
A 314.16mm
b) Young’s Modulus, (consider a load which is within the elastic limit)
P
60kN
2

190.98
314
.
16
mm
A
N
E 



190980
0.20mm
mm2
 L
110 3
L
 190980 MPa
 190.98GPa
200mm
46
Illustrative Problem
Example 2
c) % elongation,
% elongation =
Final length at fracture – original length
Original length
50

100  25%
200
Original c/s area -Final c/s area at fracture
d) % reduction in area =
Original c/s area

2


16
314.16 
314.16
4 100  36%
47
Illustrative Problem
Example 2
e) Ultimate strength,
Ultimate strength =
Original c/s area

180kN
2

572
.
96
N
/
mm
314.16mm2
( MPa)
f) Nominal breaking Strength
=
Original c/s area

160kN
 509.29MPa
314.16
g) True breaking Strength
=
c/s area at fracture
160kN

 795.38MPa
2
201.06mm
48
Illustrative Problem
Example 3
A composite bar consists of an aluminum section rigidly fastened
between a bronze section and a steel section as shown in figure.
Axial loads are applied at the positions indicated. Determine the
change in each section and the change in total length. Given
Ebr = 100GPa, Eal = 70GPa, Est = 200GPa
4kN
Bronze
A= 120 mm2
300mm
Aluminum
A= 180 mm2
13kN
400mm
Steel
A= 160 mm2
2kN
7kN
500mm
49
Illustrative Problem
Example 3
From the Example 1, we know that,
Pbr = +4kN (Tension)
Pal = -9kN (Compression)
Deformation
due
to
compressive
force
is
shortening in length, and
is considered as -ve
Pst = -7kN (Compression)
PL
stress (σ)
= 
E =
strain (ε)
AL
Change in length =
PL
L 
AE
Change in length of
bronze
= Lbr 
4000 N  300mm
120mm2 100 103 ( N / mm2 )
= 0.1mm
50
Illustrative Problem
Example 3
Change in length of
aluminum section =
 9000 N  400mm
Lal 
180mm2  70 103 ( N / mm2 )
= -0.286mm
Change in length of
steel section
=
 7000 N  500mm
Lst 
160mm2  200 103 ( N / mm2 )
= -0.109mm
Change in total
length
= Lbr  Lal  Lst  +0.1 – 0.286 - 0.109
= -0.295mm
51
Illustrative Problem
Example 4
An aluminum rod is fastened to a steel rod as shown. Axial
loads are applied at the positions shown. The area of cross
section of aluminum and steel rods are 600mm2 and 300mm2
respectively. Find maximum value of P that will satisfy the
following conditions.
a)σst ≤ 140 MPa
Take Eal = 70GPa, Est = 200GPa
b)σal ≤ 80 MPa
c)Total elongation ≤ 1mm,
2P
Aluminum
0.8m
4P
2P
Steel
2.8m
52
Illustrative Problem
solution
To find P, based on the condition, σst ≤ 140 MPa
Stress in steel must be less than or equal to 140MPa.
Hence, σst =
= 140MPa 
P
2P
Pst 2 P

 140 N / mm2
Ast Ast
140  Ast
 21000 N  21kN
2
Aluminum
4P
2P
2P
Steel
4P
2P
2P 2P
Tensile
53
Illustrative Problem
solution
To find P, based on the condition, σal ≤ 80 MPa
Stress in aluminum must be less than or equal to 80MPa.
Hence, σal =
2P
Pal

 80 N / mm2

Aal
Aal
80  Aal
P
 24000 N  24kN
2
= 80MPa
2P
Aluminum
4P
Steel
2P
4P 2P
2P
2P Compressive
2P
54
Illustrative Problem
Solution
To find P, based on the condition, total elongation ≤ 1mm
Total elongation = elongation in aluminum + elongation in steel.
1mm
 PL   PL 

 

AE
AE

 al 
 st
1mm
  2 PLal    2 PLst 
  

 
 Aal Eal   Ast Est 
1mm
  2 P  800    2 P  2800 


3  
3 
 600  70 10   300  200 10 
P = 18.1kN
Ans: P = 18.1kN (minimum of the three values)
Example:
Shearing Stress
71
• Forces P and P‘ are applied transversely
to the member AB.
• Corresponding internal forces act in the
plane of section C and are called shearing
forces.
• The resultant of the internal shear force
distribution is defined as the shear of the
section and is equal to the load P.
• The corresponding average shear stress is,
P
 ave 
A
• The shear stress distribution cannot be
assumed to be uniform.
72
Double shear
Consider the simple riveted lap joint shown in the Fig.(a). When
load is applied to the plates as shown in the figure the rivet is subjected
to shear forces tending to shear it on one plane as indicated.
But the joint with two cover plates, shown in Fig.(b), the rivet is
subjected to possible shearing on two faces, which is called as double
shear. In such cases twice the area of the rivet is resisting the applied
forces so that the shear stress set up is given by
Shear stress τ (in double shear) = P/2A
P
P
P
Fig. a
P
Fig. b
73
Examples
Single Shear
Double Shear
P F
 ave  
A A
P F
 ave  
A 2A
74
example: Shearing Stress
Pin Shearing
Stresses
To find the shearing stress in pin.
Rod BC
• The cross-sectional area for
pins at D
2
 25 mm 
6 2
A r 

491

10
m

 2 
2
• The force on the pin at C is
equal to the force exerted by
the rod BC,
 C ,ave
P
50 10 N
 
 102 MPa
6
2
A 49110 m
3
75
Example: Double shear
Rod AB
• The cross-sectional area for pins at D,
&E
2
 25 mm 
6
2
A r 
  49110 m
 2 
2
• The pin at A is in double shear with a
total force equal to the force exerted
by the boom AB,
P
40 kN
 A,ave 

 40.7 MPa
6
2
2 A 2(49110 m )
85
Poisson’s Ratio = µ
For most engineering metals the value of µ lies between 0.25 and 0.33
y
In general
Lz
Ly
P
P
x
Lx
z
 l y
Poisson’s Ratio =
Lateral strain
Strain in the direction of

l x
ly
lx
 l z
OR  l
x
lz
lx
86
Poisson’s Ratio = µ
In general
y
Ly
Lz
Px
Px
x
Lx
z
Strain in X-direction = εx
Strain in Y-direction = εy
l x

lx
Strain in Z-direction = εz


l z
lz

l x
lx
l y
ly

l x
lx
87
Poisson’s Ratio
y
Py
Lz
Ly
x
Lx
z
Py
 l x
Poisson’s Ratio =
Lateral strain

Strain in the direction of
Strain in X-direction = εx

l x
lx

l y
lx
ly
l y
ly
 l z
OR  l
y
lz
ly
88
Poisson’s Ratio
y
Pz
Lz
Ly
x
Lx
z
Pz
 l x
Poisson’s Ratio =
Lateral strain

Strain in the direction of
Strain in X-direction = εx

l x
lx
l z

 l y
lx
lz
l z
lz
OR

l z
ly
lz
```