MECHANICAL PROPERTIES OF SOLIDS

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.

𝐒𝐭𝐫𝐞𝐬𝐬 = 𝑭/𝑨

Stress is a scalar quantity. Its SI unit is 𝑁𝑚⁻² or 𝑝𝑎𝑠𝑐𝑎𝑙 (𝑃𝑎). Its dimensional formula is [𝑀𝐿⁻¹𝑇⁻²]

Strain

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

𝐒𝐭𝐫𝐚𝐢𝐧 = 𝒄𝒉𝒂𝒏𝒈𝒆 𝒊𝒏 𝒄𝒐𝒏𝒇𝒊𝒈𝒖𝒓𝒂𝒕𝒊𝒐𝒏/𝒐𝒓𝒊𝒈𝒊𝒏𝒂𝒍 𝒄𝒐𝒏𝒇𝒊𝒈𝒖𝒓𝒂𝒕𝒊𝒐𝒏

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 (𝜎)	(1) Longitudinal strain (𝜀)
(2) Tangential stress / Shearing stress (𝜎𝑆)	(2) Shearing strain (𝜃)
(3) Hydraulic stress / Volume stress / Bulk stress (𝑃)	(3) Volume strain

Normal stress(𝝈)

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.

𝐿𝑜𝑛𝑔𝑖𝑡𝑢𝑑𝑖𝑛𝑎𝑙 𝑠𝑡𝑟𝑎𝑖𝑛 = 𝐶ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑙𝑒𝑛𝑔𝑡ℎ/𝑂𝑟𝑖𝑔𝑖𝑛𝑎𝑙 𝑙𝑒𝑛𝑔𝑡ℎ

𝜺 = 𝚫𝑳/𝑳

Tangential stress / Shearing stress(𝝈_𝑺)

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(𝜽)

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 = Δ𝑥/𝐿 tan 𝜃 = Δ𝑥/𝐿 If 𝜃 is small, 𝑡𝑎𝑛 𝜃 ≈ 𝜃 then, 𝜽 = 𝚫𝒙/𝑳

Hydraulic stress / Volume stress / Bulk stress(𝒑)

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.

𝑽𝒐𝒍𝒖𝒎𝒆 𝒔𝒕𝒓𝒂𝒊𝒏 = − ∆𝑽/𝑽

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

Hooke’s law

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

𝑆𝑡𝑟𝑒𝑠𝑠 ∝ 𝑆𝑡𝑟𝑎𝑖𝑛

𝑆𝑡𝑟𝑒𝑠𝑠 = 𝑘 (𝑆𝑡𝑟𝑎𝑖𝑛)

The constant 𝑘 is known as modulus of elasticity. Its unit is 𝑁𝑚⁻² and dimensions are [𝑀𝐿⁻¹𝑇⁻²]

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 𝑂 and 𝐴, the curve is linear, and Hooke’s law is obeyed. The body regains its original dimensions, and the body behaves as an elastic body. In the region from 𝐴 to 𝐵, stress and strain are not proportional, but the body still regains its original dimensions after the removal of load. The point 𝐵 is called Yield point (elastic limit) and the corresponding stress is called Yield strength (𝜎_𝑦). If the stress is increased beyond 𝐵, the strain increases rapidly. This is represented by the region between 𝐵 and 𝐷. When the load is removed, say at 𝐶, 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 𝐷 on graph is the ultimate tensile strength(𝜎_𝑢) of the material. Beyond this point, additional strain is produced even by a reduced applied force and fracture occurs at 𝐸. 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 Hooke’s 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) Young’s modulus (𝑌) (ii) Shear modulus / Rigidity modulus (𝐺) (iii) Bulk modulus (𝐵)

Young’s modulus(𝒀)

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

𝑌 = 𝑁𝑜𝑟𝑚𝑎𝑙 𝑠𝑡𝑟𝑒𝑠𝑠/𝑙𝑜𝑛𝑔𝑖𝑡𝑢𝑑𝑖𝑛𝑎𝑙 𝑠𝑡𝑟𝑎𝑖𝑛

𝑌 = 𝜎/𝜀

𝒀 = (𝑭/𝑨)/(∆𝑳/𝑳) = 𝑭𝑳/𝑨∆𝑳

The unit of young’s modulus is 𝑁𝑚⁻². Generally, for metals Young’s moduli are large; Hence they are more elastic in nature.

Determination of Young’s 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 𝐴 is called Reference wire carries a main scale and a pan to place a weight. The wire 𝐵 is called experimental wire also carries a pan in which known weights can be placed. A vernier scale 𝑉 is attached to a pointer at the bottom of the wire 𝐵 and main scale is fixed to wire 𝐴.

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 Young’s modulus of material of a wire

If 𝑟 and 𝐿 are the initial radius and length of the wire 𝐵 respectively. The area of cross-section is 𝜋𝑟². Let ∆𝐿 be the elongation produced by the mass 𝑀, 𝑌𝑜𝑢𝑛𝑔’𝑠 𝑚𝑜𝑑𝑢𝑙𝑢𝑠, 𝑌 = 𝜎/𝜀 = 𝐹𝐿/𝐴∆𝐿

𝑌 = (𝑀𝑔)𝐿/(𝜋𝑟²)∆𝐿

Shearing modulus/Rigidity modulus(𝑮)

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

𝐺 = 𝑆ℎ𝑒𝑎𝑟𝑖𝑛𝑔 𝑠𝑡𝑟𝑒𝑠𝑠/𝑆ℎ𝑒𝑎𝑟𝑖𝑛𝑔 𝑠𝑡𝑟𝑎𝑖𝑛

𝐺 = 𝜎_𝑠/𝜃

𝑮 = (𝑭/𝑨)/(∆𝒙/𝑳) = 𝑭𝑳/𝑨∆𝒙

Its SI unit is 𝑁𝑚⁻². It can be seen that Shear modulus is generally less than Young’s modulus for most materials and ≈ 𝑌/3.

Bulk modulus(𝑩)

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

𝑩 = 𝑯𝒚𝒅𝒓𝒂𝒖𝒍𝒍𝒊𝒄 𝒔𝒕𝒓𝒆𝒔𝒔/𝑯𝒚𝒅𝒓𝒂𝒖𝒍𝒍𝒊𝒄 𝒔𝒕𝒓𝒂𝒊𝒏 = −𝒑/(∆𝑽/𝑽) = −𝒑𝑽/∆𝑽

Negative sign shows that increase in pressure (p) causes decrease in volume (𝑉). Its SI unit is 𝑁𝑚⁻².

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(𝒌)

The reciprocal of the bulk modulus is called Compressibility.

𝑘 = 1/𝐵 = 1/[− 𝑝(∆𝑉/𝑉)]

𝒌 = −∆𝑽/𝒑𝑽

𝑘 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.

𝐿𝑎𝑡𝑒𝑟𝑎𝑙 𝑠𝑡𝑟𝑎𝑖𝑛 ∝ 𝐿𝑜𝑛𝑔𝑖𝑡𝑢𝑑𝑖𝑛𝑎𝑙 𝑠𝑡𝑟𝑎𝑖𝑛

Poisons ratio(𝝈)

The ratio of the lateral strain to the longitudinal strain is called Poison’s ratio.

𝑃𝑜𝑖𝑠𝑜𝑛’𝑠 𝑟𝑎𝑡𝑖𝑜 = 𝐿𝑎𝑡𝑒𝑟𝑎𝑙 𝑠𝑡𝑟𝑎𝑖𝑛/𝐿𝑜𝑛𝑔𝑖𝑡𝑢𝑑𝑖𝑛𝑎𝑙 𝑠𝑡𝑟𝑎𝑖𝑛

𝑷𝒐𝒊𝒔𝒐𝒏’𝒔 𝒓𝒂𝒕𝒊𝒐 = (∆𝒅/𝒅)/(∆𝑳/𝑳) = 𝑳∆𝒅/𝒅∆𝑳

Poison’s ratio is a ratio of two strains. It is a pure number and has no dimensions and unit. The practical value of Poison’s 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 𝐿 and area of cross-section 𝐴. Let a force 𝐹 be applied to stretch the wire. If 𝑙 be the length through which the wire is stretched. Then, 𝑌 = (𝐹/𝐴)/(𝑙/𝐿) = 𝐹𝐿/𝐴𝑙

𝐹 = 𝑌𝐴𝑙/𝐿

If the wire is stretched through a length 𝑑𝑙, work done is given by, 𝑑

𝑊 = 𝐹𝑑𝑙 𝑑𝑊 = (𝑌𝐴𝑙/𝐿)𝑑𝑙

Total work done to stretch the wire from 0 to 𝑙 is, 𝑊 = ∫𝑑𝑊 = ₀∫^l(𝑌𝐴𝑙/𝐿)𝑑𝑙

𝑊 = 𝑌𝐴𝑙²/2𝐿

𝑊 = (1/2)𝑌(𝑙/𝐿)²𝐴𝐿

𝑊 = (1/2)𝜎𝜀(𝜀)²𝐴𝐿

𝑊 = (1/2)𝜎𝜀 (𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑤𝑖𝑟𝑒)

𝑼 = (𝟏/𝟐)𝝈𝜺 (𝑽𝒐𝒍𝒖𝒎𝒆 𝒐𝒇 𝒕𝒉𝒆 𝒘𝒊𝒓𝒆)

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

𝒖 = (𝟏/𝟐)𝝈𝜺

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.

𝐄𝐱𝐩𝐥𝐚𝐧𝐚𝐭𝐢𝐨𝐧: 𝑆𝑡𝑟𝑒𝑠𝑠 = 𝐿𝑜𝑎𝑑/𝐴𝑟𝑒𝑎 = 𝑀𝑔/𝜋𝑟² where 𝑟 → radius of the rope required

Let us take the lifted mass 𝑀 = 10⁵𝑘𝑔 and 𝑔 = 10𝑚𝑠⁻². The elastic limit of the steel is 30 × 10⁷𝑁𝑚², Then the maximum stress on the rope is 30 × 10⁷𝑁𝑚². Radius, 𝑟 = [(10⁵ × 10)/(3.14 × 30 × 10⁷)]^1/2 = 0.0325𝑚 = 3.25𝑐𝑚

The radius of the steel rope to lift 10⁵𝑘𝑔 should be about 3𝑐𝑚. 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 (𝑙), breadth (𝑏) and depth (𝑑) loaded weight is as shown. The depression of the beam is,

𝛿 = 𝑀𝑔𝑙³/4𝑏𝑑³𝑌

𝛿 ∝ 1/𝑑³

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

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 ℎ is the height of the mountain and 𝜌 be the density of the rocks of the mountain, then the pressure at the base of the mountain=𝜌𝑔ℎ = 𝑠𝑡𝑟𝑒𝑠𝑠. The elastic limit of a typical rock = 3 × 10⁸𝑁𝑚⁻². The stress must be less than the elastic limit; otherwise the rock begins to sink under its own weight.

𝜌𝑔ℎ < 3 × 10⁸

ℎ < (3 × 10⁸)/𝜌𝑔 = (3 × 10⁸)/(3 × 10³ × 10)

ℎ < 10⁴𝑚

ℎ ≈ 10𝑘𝑚

It may be noted that the height of the Mount Everest is nearly 9𝑘𝑚.

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 Poison’s ratio.

Two Marks.

1. State and explain Hooke’s law of elasticity.

2. What are elastomers? Give example.

3. Determine the volume contraction of the solid copper tube, 10 𝑐𝑚 on its edge, when subjected to hydraulic pressure of 7.0 × 10⁶𝑃𝑎. (Given bulk modulus of copper = 140 × 10 𝑁𝑚⁻²).

Three Marks.

1. Define three types of stress.

2. Define three types of strains.

3. Define stress, strain and elastic limit.

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

5. Arrive at an expression for Young’s modulus in case of stretched string.

6. A steel rod of radius 0.01 𝑚 and length 2 𝑚. A 100 𝑁 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 Young’s modulus of a material wire.

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

Five Marks.

1. State and explain Hooke’s law. Draw stress-strain curve and label the parts.

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