1.
Lecture 1
Stresses & Strains
Structure II (AR-106 G)
BY- AR. KHURRAM ALI
Asst. Professor
Gateway College of architecture
Sonipat, Haryana
2.
Introduction
Sessional Works:
• Please attend all scheduled lectures, seminars and
practicals
• Please note: attendance falling below 75% may lead to failing
the course irrespective of the academic performance.
Attendance:
All assignments to be submitted within the time frame
given on the same day.
The absentees will have to submit the assignment
before the next class and there will be a deduction of 50
% marks in the assignments of the absentees, any
submission beyond this time frame will not be accepted
3.
Classroom tests:
• Two test will be conducted in the semester carrying a
weightage of 50 % of the sessional marks
Introduction
Course Evaluation:
You’re required to achieve a minimum of 50% in each
classroom test to pass the course.
•Sessional Marks: 50
•Theory Exam : 50
•Total Marks : 100
4.
• Whenever a body is
subjected to an external
force, it tends to undergo
deformation (i.e. change
in shape or dimensions).
• Due to cohesion between
the molecules, the body
resists deformation. This
internal resistance is
equal and opposite to the
applied force.
Introduction
5.
Stress
• The internal resistance per unit area of cross section
offered by a body against the deformation is called stress.
• Mathematically stress may be defined as the force per unit
area i.e.
stress, (σ) = P/A
where, P = load or force acting on the body
A = cross-sectional area of the body
Unit of Stress- Unit of stress depends upon the unit of load
or force and unit of area. In SI units, stress is expressed as
N/m2 or N/mm2.
N/m2 is called Pascal abbreviated as Pa.
6.
Types of Stresses
• The important types of stresses are:
1. Direct stress or Normal stress
2. Shear stress
3. Bending stress
4. Torsional stress
5. Thermal stress
7.
Direct stress
• Direct stress is that
which acts
perpendicular to the
cross-section of the
member.
• Direct stress is also
known as normal stress.
• Direct stress is either
tensile or compressive
in nature according to
the nature of loading.
8.
Shear stress (Tangential stress)
• When two equal and
opposite forces are
acting tangentially to
the cross-section, the
stress induced in the
member is known as
shear stress.
• Due to these equal and
opposite forces, the
member tends to shear
off across the section.
9.
Bending stress (Longitudinal stress)
• When the beam is loaded
with some external loads,
the bending moments
and shear forces set up at
all the sections of the
beam.
• The material of the beam
will offer resistance or
stresses to bending. The
stresses produced to
resist the bending
moment are known as
bending stresses.
10.
Torsional stress
• The shear stress set up
in the shaft when equal
and opposite torques
(twisting moments) are
applied to the ends of a
shaft about its axis, is
called torsional stress.
11.
Thermal stress ( temperature stress)
• When the temperature of a material is
changed, its dimensions change.
• if this change in dimensions is prevented, then
a stress is set up in the material, which is
called a thermal stress.
12.
Strain
• Strain is a measure of the deformation caused
by the loaded body.
• Mathematically, the ratio of change in
dimension of the body to the original
dimension is known as strain i.e.
strain, (e) = Change in dimension/Original dimension
Unit of strain- strain is a ratio, as such it has
no units.
13.
Types of strain
• The important types of
simple strains are-
1. Tensile strain- Figure
shows a body subjected to
an axial tensile load P,
which increases the length
of the body from l to l+ᵹl.
hence Tensile strain,
et = increase in length/
original length
= δl/l
14.
Contd.
2. Compressive strain-
Figure shows a body
subjected to an axial
compressive load P,
which decreases the
length of the body from l
to l- δl.
hence Comp. strain,
ec = decrease in length/
original length
= δl/l
15.
Contd.
3. Shear strain- It is a
measure of the angle
through which a body is
distorted under the
action of shear forces.
Shear strain, (eq) = tan φ
or φ (only)
Example- when bolt is
turned by a spanner the
material of the bolt is in
a state of shear strain.
16.
Contd.
4. Volumetric strain- The
ratio between the
change in volume and
the original volume is
known as volumetric
strain or Bulk strain.
ev = δV/V
17.
Elasticity and Elastic limit
• A material is said to be elastic if the deformation produced
in it by external forces, completely disappears on the
removal of external forces.
• This property by virtue of which certain materials return
back to their original position after the removal of the
external forces, is called Elasticity.
• There is a limiting value of force upto and within which, the
deformation completely disappears on the removal of the
force.
• The value of stress corresponding to this limiting force upto
which the material is perfectly elastic is known as Elastic
limit.
• Beyond the elastic limit, the material gets into plastic stage.
18.
Hooke’s Law
• Robert Hooke in 1676 discovered experimentally
a basic law in elasticity.
• Hooke’s law states that when a material is
loaded within elastic limit, the stress is directly
proportional to strain produced by the stress.
therefore, Stress ȣ Strain
stress = a constant x strain
or stress/strain = a constant
this constant is known as Co-efficient of elasticity
or Modulus of elasticity.
19.
Young’s Modulus of Elasticity (E)
• The ratio between tensile stress and tensile strain or
between compressive stress and compressive strain
is called Young’s Modulus of elasticity.
hence, E = σt/et or σc/ec
the unit of Young’s Modulus of elasticity are N/m2 or
N/mm2
20.
Illustrative examples
1. Find the minimum diameter of a steel wire which is used to raise a
load of 5000N if the stress in the wire is not to exceed 95 MN/m2
Solution- Given, Load (P) = 5000N
Stress (σ) = 95 MN/m2 = 95x106 N/m2
= 95 N/mm2
now, let diameter of wire = d
Cross-sectional area of wire, A = πd2/4
Stress, σ = P/A
95 = 5000x4/πd2
d2 = 67.01
d = 8.19mm
21.
Contd.
2. A wooden tie is 75mm wide, 150mm deep and 2.5m long. It is subjected
to an axial pull of 40kN. The extension of the member is found to be
0.80mm. Find the young’s modulus of elasticity for the tie material.
Solution- Area of tie, A = 75x150 = 11250mm2
Length of the tie, l = 2.5m = 2500mm
Axial pull, P = 40kN = 40000N
Extension, δl = 0.80mm
Stress, σ = P/A = 40000/11250
= 3.56 N/mm2
Strain, e = δl/l = 0.80/2500
= 0.00032
Young’s Modulus, E = σ/e = 3.56/0.00032
= 11125 N/mm2 = 11.125 kN/mm2
22.
Test
1. A rod 16mm in diameter
and 2.5m long is
subjected to an axial pull
of 25kN. If the Young’s
modulus of elasticity of
the material of the rod is
210 MN/mm2;
determine:
i) The stress
ii) The strain
iii) The elongation of the
rod
1. A rod 18mm in diameter
and 2.4m long is
subjected to an axial pull
of 20kN. If the Young’s
modulus of elasticity of
the material of the rod is
210 MN/mm2;
determine:
i) The stress
ii) The strain
iii) The elongation of the
rod
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