Lecture 1

Stresses & Strains

Structure II (AR-106 G)

BY- AR. KHURRAM ALI Asst. Professor Gateway College of architecture Sonipat, Haryana

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

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

• 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

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.

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

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.

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.

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.

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.

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.

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.

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

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

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.

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

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.

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.

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

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

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

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