**This article is formulated according to the 9th chapter in the NCERT Syllabus of 11th Std Physics OR 1st PUC Physics.**

## Elasticity

The property of a body, due to which it tends to regain its original size and shape when deforming force is removed, is called elasticity. The deformation caused is called as elastic deformation.

## Model of Spring-ball system

Elastic behaviour of the solids can be explained by the model of Spring-ball system. The ball represents atoms and springs represent interatomic forces. When the solid is deformed, the atoms are displaced from their mean position causing change in inter atomic distances. When the deforming force is removed the inter-atomic forces tend to drive them back to their original positions and the body gains the original shape.

## Plasticity

The property of a body due to which it does not regain its original size and shape when the deforming force is removed is called plasticity. The substances are called as plastics.

## Stress

When a deforming force is applied on a body, the restoring force is developed inside a body. The restoring force per unit area is known as stress. Stress can also be defined as deforming force per unit area, because magnitude of deforming force and restoring force are equal.

๐๐ญ๐ซ๐๐ฌ๐ฌ = ๐ญ/๐จ

Stress is a scalar quantity. Its SI unit is ๐๐^{โ2} or ๐๐๐ ๐๐๐ (๐๐). Its dimensional formula is [๐๐ฟ^{โ1}๐^{โ2}]

## Strain

The ratio of Change in configuration to the original configuration of the body is called strain.

๐๐ญ๐ซ๐๐ข๐ง = ๐๐๐๐๐๐ ๐๐ ๐๐๐๐๐๐๐๐๐๐๐๐๐/๐๐๐๐๐๐๐๐ ๐๐๐๐๐๐๐๐๐๐๐๐๐

Strain has no unit. It is dimensionless quantity. It is a ratio.

## Types of Stress and strain

There are three types in each.

Stress | Strain |

(1) Normal stress (๐) | (1) Longitudinal strain (๐) |

(2) Tangential stress / Shearing stress (๐๐) | (2) Shearing strain (๐) |

(3) Hydraulic stress / Volume stress / Bulk stress (๐) | (3) Volume strain |

## Normal stress(๐)

It is defined as the restoring force per unit area perpendicular to the surface of the body. Normal stress is of two types.

(i) Tensile stress

(ii) Compressive stress

## Tensile stress

When two equal and opposite forces are applied at the ends of a circular rod to increase its length, a restoring force normal to the cross-sectional area of the rod is developed. This restoring force per unit area of cross-section is known as tensile stress.

## Compressive stress

When two equal and opposite forces are applied at the ends of a circular rod to decrease its length, a restoring force normal to the cross-sectional area of the rod is developed. This restoring force per unit area of cross-section is known as compressive stress.

## Longitudinal strain

This type of strain is produced when the body is under the tensile or compressive stress. It is defined as the ratio of the change in length to the original length of the body.

๐ฟ๐๐๐๐๐ก๐ข๐๐๐๐๐ ๐ ๐ก๐๐๐๐ = ๐ถโ๐๐๐๐ ๐๐ ๐๐๐๐๐กโ/๐๐๐๐๐๐๐๐ ๐๐๐๐๐กโ

๐บ = ๐ซ๐ณ/๐ณ

## Tangential stress / Shearing stress(๐_{๐บ})

The restoring force per unit area developed due to the applied tangential force is known as tangential stress or Shearing stress. When two equal and opposite forces act along the tangent to the surfaces of the opposite faces of an object then one face of the object is displaced with respect to the other face.

## Shearing strain(๐ฝ)

This type of strain is produced when the body is under tangential of shearing stress. It is defined as the angle through which the face of the body originally perpendicular to the fixed face is turned when it is under shearing stress. Shearing strain = ฮ๐ฅ/๐ฟ tan ๐ = ฮ๐ฅ/๐ฟ If ๐ is small, ๐ก๐๐ ๐ โ ๐ then, ๐ฝ = ๐ซ๐/๐ณ

## Hydraulic stress / Volume stress / Bulk stress(๐)

When an object is immersed in a fluid, the fluid exerts force on the surfaces of the object as a result the volume of the object decreases,and the object is under stress known as hydraulic stress. During hydraulic stress there is no change in the geometrical shape of the object.

## Volume strain

This type of strain is produced when the body is under the hydraulic stress. It is defined as the ratio of change in volume to the original volume.

๐ฝ๐๐๐๐๐ ๐๐๐๐๐๐ = โ โ๐ฝ/๐ฝ

Negative sign indicates that volume decreases when the body is under bulk stress.

## Hookeโs law

For small deformations, the stress and strain are proportional to each other.

๐๐ก๐๐๐ ๐ โ ๐๐ก๐๐๐๐

๐๐ก๐๐๐ ๐ = ๐ (๐๐ก๐๐๐๐)

The constant ๐ is known as modulus of elasticity. Its unit is ๐๐^{โ2} and dimensions are [๐๐ฟ^{โ1}๐^{โ2}]

## Stress-Strain Curve

The stress-strain curve changes from material to material. This curve helps us to understand how given materials deform with increasing loads.

**Explanation:** In the region between ๐ and ๐ด, the curve is linear, and Hookeโs law is obeyed. The body regains its original dimensions, and the body behaves as an elastic body. In the region from ๐ด to ๐ต, stress and strain are not proportional, but the body still regains its original dimensions after the removal of load. The point ๐ต is called Yield point (elastic limit) and the corresponding stress is called Yield strength (๐_{๐ฆ}). If the stress is increased beyond ๐ต, the strain increases rapidly. This is represented by the region between ๐ต and ๐ท. When the load is removed, say at ๐ถ, the body does not regain its original dimensions. The material is said to have a permanent set. The deformation is said to be plastic deformation. The point ๐ท on graph is the ultimate tensile strength(๐_{๐ข}) of the material. Beyond this point, additional strain is produced even by a reduced applied force and fracture occurs at ๐ธ. If the ultimate strength and fracture points are close, the material is said to be brittle. If they are far apart, the material is said to be ductile.

## Elastomers

The materials having large elastic region but does not obey Hookeโs law and have no well-defined plastic region are called elastomers. Ex: Rubber, tissue of aorta etc.

## Elastic moduli

The ratio of stress and strain is called modulus of elasticity.

## Types of moduli of elasticity

There are three types (i) Youngโs modulus (๐) (ii) Shear modulus / Rigidity modulus (๐บ) (iii) Bulk modulus (๐ต)

## Youngโs modulus(๐)

The ratio of Normal (tensile or compressive) stress to the longitudinal strain is defined as youngโs modulus.

๐ = ๐๐๐๐๐๐ ๐ ๐ก๐๐๐ ๐ /๐๐๐๐๐๐ก๐ข๐๐๐๐๐ ๐ ๐ก๐๐๐๐

๐ = ๐/๐

๐ = (๐ญ/๐จ)/(โ๐ณ/๐ณ) = ๐ญ๐ณ/๐จโ๐ณ

The unit of youngโs modulus is ๐๐^{โ2}. Generally, for metals Youngโs moduli are large; Hence they are more elastic in nature.

## Determination of Youngโs modulus of material of a wire

The arrangement consists of two long straight wires of same length and equal radius suspended side by side from a fixed rigid support. The wire ๐ด is called Reference wire carries a main scale and a pan to place a weight. The wire ๐ต is called experimental wire also carries a pan in which known weights can be placed. A vernier scale ๐ is attached to a pointer at the bottom of the wire ๐ต and main scale is fixed to wire ๐ด.

Both the wires are given an initial small load to keep them straight. The vernier reading is noted now. The experimental wire is gradually loaded with more weights to bring it under tensile stress and the vernier reading is noted again. The difference between two vernier readings gives the elongation produced in the wire.

## Expression for Youngโs modulus of material of a wire

If ๐ and ๐ฟ are the initial radius and length of the wire ๐ต respectively. The area of cross-section is ๐๐^{2}. Let โ๐ฟ be the elongation produced by the mass ๐, ๐๐๐ข๐๐โ๐ ๐๐๐๐ข๐๐ข๐ , ๐ = ๐/๐ = ๐น๐ฟ/๐ดโ๐ฟ

๐ = (๐๐)๐ฟ/(๐๐^{2})โ๐ฟ

## Shearing modulus/Rigidity modulus(๐ฎ)

The ratio of shearing stress to the corresponding shearing strain is called shear modulus.

๐บ = ๐โ๐๐๐๐๐๐ ๐ ๐ก๐๐๐ ๐ /๐โ๐๐๐๐๐๐ ๐ ๐ก๐๐๐๐

๐บ = ๐_{๐ }/๐

๐ฎ = (๐ญ/๐จ)/(โ๐/๐ณ) = ๐ญ๐ณ/๐จโ๐

Its SI unit is ๐๐^{โ2}. It can be seen that Shear modulus is generally less than Youngโs modulus for most materials and โ ๐/3.

## Bulk modulus(๐ฉ)

The ratio of hydraulic stress to the corresponding hydraulic strain (Volume strain) is called bulk modulus.

๐ฉ = ๐ฏ๐๐ ๐๐๐๐๐๐๐ ๐๐๐๐๐๐/๐ฏ๐๐ ๐๐๐๐๐๐๐ ๐๐๐๐๐๐ = โ๐/(โ๐ฝ/๐ฝ) = โ๐๐ฝ/โ๐ฝ

Negative sign shows that increase in pressure (p) causes decrease in volume (๐). Its SI unit is ๐๐^{โ2}.

**Note:** Bulk modulus for a perfect rigid body and ideal gas is infinite. A solid has all types of moduli of elasticity but fluids have only bulk modulus of elasticity.

## Compressibility(๐)

The reciprocal of the bulk modulus is called Compressibility.

๐ = 1/๐ต = 1/[โ ๐(โ๐/๐)]

๐ = โโ๐ฝ/๐๐ฝ

๐ is also defined as the fractional change in volume per unit increase in pressure. Solids are least compressible than liquids. Gases are more compressible than liquids and solids.

## Lateral strain

The ratio of change in diameter to the original diameter is called lateral strain. Within elastic limit this lateral strain is directly proportional to the longitudinal strain.

๐ฟ๐๐ก๐๐๐๐ ๐ ๐ก๐๐๐๐ โ ๐ฟ๐๐๐๐๐ก๐ข๐๐๐๐๐ ๐ ๐ก๐๐๐๐

## Poisons ratio(๐)

The ratio of the lateral strain to the longitudinal strain is called Poisonโs ratio.

๐๐๐๐ ๐๐โ๐ ๐๐๐ก๐๐ = ๐ฟ๐๐ก๐๐๐๐ ๐ ๐ก๐๐๐๐/๐ฟ๐๐๐๐๐ก๐ข๐๐๐๐๐ ๐ ๐ก๐๐๐๐

๐ท๐๐๐๐๐โ๐ ๐๐๐๐๐ = (โ๐ /๐ )/(โ๐ณ/๐ณ) = ๐ณโ๐ /๐ โ๐ณ

Poisonโs ratio is a ratio of two strains. It is a pure number and has no dimensions and unit. The practical value of Poisonโs ratio lies between 0 and 0.5.

## Elastic Potential energy

When a wire is put under a tensile stress, work is done against the inter-atomic forces. This work is stored in the wire in the form of Elastic potential energy.

## Expression for Elastic Potential energy

Consider a wire of length ๐ฟ and area of cross-section ๐ด. Let a force ๐น be applied to stretch the wire. If ๐ be the length through which the wire is stretched. Then, ๐ = (๐น/๐ด)/(๐/๐ฟ) = ๐น๐ฟ/๐ด๐

๐น = ๐๐ด๐/๐ฟ

If the wire is stretched through a length ๐๐, work done is given by, ๐

๐ = ๐น๐๐ ๐๐ = (๐๐ด๐/๐ฟ)๐๐

Total work done to stretch the wire from 0 to ๐ is, ๐ = โซ๐๐ = _{0}โซ^{l}(๐๐ด๐/๐ฟ)๐๐

๐ = ๐๐ด๐^{2}/2๐ฟ

๐ = (1/2)๐(๐/๐ฟ)^{2}๐ด๐ฟ

๐ = (1/2)๐๐(๐)^{2}๐ด๐ฟ

๐ = (1/2)๐๐ (๐๐๐๐ข๐๐ ๐๐ ๐กโ๐ ๐ค๐๐๐)

๐ผ = (๐/๐)๐๐บ (๐ฝ๐๐๐๐๐ ๐๐ ๐๐๐ ๐๐๐๐)

The elastic potential energy per unit volume of the wire (๐ข) is given by,

๐ = (๐/๐)๐๐บ

## Applications of Elastic behaviour of materials

1) A Crane is used for lifting and moving heavy loads from one place to another. The crane makes use of a thick metallic rope. The maximum load lifted should be such that, the elastic limit of the material of the rope is not exceeded.

๐๐ฑ๐ฉ๐ฅ๐๐ง๐๐ญ๐ข๐จ๐ง: ๐๐ก๐๐๐ ๐ = ๐ฟ๐๐๐/๐ด๐๐๐ = ๐๐/๐๐^{2} where ๐ โ radius of the rope required

Let us take the lifted mass ๐ = 10^{5}๐๐ and ๐ = 10๐๐ ^{โ2}. The elastic limit of the steel is 30 ร 10^{7}๐๐^{2}, Then the maximum stress on the rope is 30 ร 10^{7}๐๐^{2}. Radius, ๐ = [(10^{5} ร 10)/(3.14 ร 30 ร 10^{7})]^{1/2} = 0.0325๐ = 3.25๐๐

The radius of the steel rope to lift 10^{5}๐๐ should be about 3๐๐. Practically it will be a rigid rod, but the rope is made up of a large number of thin wires braided together to make it flexible.

2) Bending of beam

A bridge has to design such that it can withstand the load of the flowing traffic, the force of wind and its own weight.

**Explanation:** When a beam of length (๐), breadth (๐) and depth (๐) loaded weight is as shown. The depression of the beam is,

๐ฟ = ๐๐๐^{3}/4๐๐^{3}๐

๐ฟ โ 1/๐^{3}

The depression in the loaded beam can be decreased effectively by increasing ๐.

3) In building and bridge construction, Iron girders used are not rectangular shape but in the shape of letter I. The I Shaped girder is much stronger than the rectangular shaped girder.

4) To estimate the maximum height of the mountain.

**Explanation:** The stress due to all the material on the top of the mountain should be less than the critical shearing stress at which the rocks flow. If โ is the height of the mountain and ๐ be the density of the rocks of the mountain, then the pressure at the base of the mountain=๐๐โ = ๐ ๐ก๐๐๐ ๐ . The elastic limit of a typical rock = 3 ร 10^{8}๐๐^{โ2}. The stress must be less than the elastic limit; otherwise the rock begins to sink under its own weight.

๐๐โ < 3 ร 10^{8}

โ < (3 ร 10^{8})/๐๐ = (3 ร 10^{8})/(3 ร 10^{3} ร 10)

โ < 10^{4}๐

โ โ 10๐๐

It may be noted that the height of the Mount Everest is nearly 9๐๐.

## Important Questions for Exam

**One Mark.**

1. What is mean by elasticity?

2. Is stress a scalar or a vector?

3. Mention the SI unit of stress.

4. Among steel and rubber which one is more elastic?

5. Define lateral strain.

6. Define Poisonโs ratio.

**Two Marks.**

1. State and explain Hookeโs law of elasticity.

2. What are elastomers? Give example.

3. Determine the volume contraction of the solid copper tube, 10 ๐๐ on its edge, when subjected to hydraulic pressure of 7.0 ร 10^{6}๐๐. (Given bulk modulus of copper = 140 ร 10 ๐๐^{โ2}).

**Three Marks.**

1. Define three types of stress.

2. Define three types of strains.

3. Define stress, strain and elastic limit.

4. Define Youngโs modulus of the material of a wire. Give an expression for it. Mention its SI unit.

5. Arrive at an expression for Youngโs modulus in case of stretched string.

6. A steel rod of radius 0.01 ๐ and length 2 ๐. A 100 ๐ of force stretches it along its length. Calculate the (a) stress (b) elongation.

7. Mention three types of moduli of elasticity. or Define different types of stress and strain modulus.

8. What is mean by plasticity? Define Youngโs modulus of a material wire.

9. Draw a typical stress-strain curve and label yield point, fracture point.

**Five Marks.**

1. State and explain Hookeโs law. Draw stress-strain curve and label the parts.

2. What is elastic potential energy? Obtain the expression for elastic potential energy.