"Elastic Properties of Solids"

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Why does a spring stretch when pulled and return to its original shape when released? Why do some materials break while others bend? This chapter explores the elastic behaviour of solids — how materials deform under force and recover when the force is removed. You will learn about stress, strain, Hooke's law, different elastic moduli, and the stress-strain curve that characterises a material's mechanical properties.

Key Concepts

8.1 Elasticity and Plasticity

Elasticity: The property of a material by virtue of which it regains its original shape and size after the removal of the deforming force.

Plasticity: The property by which a material retains its deformed shape even after the removal of the deforming force.

Elastic limit: The maximum stress a material can withstand without permanent deformation. Beyond this limit, the material behaves plastically.

8.2 Inter-Atomic Forces and Deformation

At equilibrium separation $R_0$, the net inter-atomic force is zero.

8.3 Stress

Stress is the restoring force per unit area developed inside a body in response to an applied deforming force.

$\text{Stress}=\frac{F}{A}$

SI unit: N/m² or pascal (Pa)

Types of stress:

8.4 Strain

Strain is the fractional change in dimension produced by stress. It is a dimensionless quantity.

8.5 Hooke's Law

For small deformations, stress is directly proportional to strain:

$\text{Stress}\propto \text{Strain}$

$\frac{\text{Stress}}{\text{Strain}}=\text{Elastic Modulus (constant)}$

Hooke's law is valid only within the elastic limit.

8.6 Elastic Moduli

Young's Modulus (Y): Measures resistance to longitudinal deformation.

$Y=\frac{\text{Longitudinal Stress}}{\text{Longitudinal Strain}}=\frac{F/A}{\Delta L/L}=\frac{FL}{A\Delta L}$

Bulk Modulus (B): Measures resistance to uniform compression.

$B=-\frac{\Delta P}{\Delta V/V}$

The negative sign indicates that an increase in pressure causes a decrease in volume.

Shear Modulus (G or η): Measures resistance to shear deformation.

$G=\frac{\text{Shear Stress}}{\text{Shear Strain}}=\frac{F/A}{\theta }$

Compressibility is the reciprocal of bulk modulus: $K=\frac{1}{B}$

8.7 Poisson's Ratio (σ)

When a wire is stretched, it becomes longer AND thinner. Poisson's ratio relates lateral strain to longitudinal strain:

$\sigma =-\frac{\text{Lateral Strain}}{\text{Longitudinal Strain}}=-\frac{\Delta d/d}{\Delta L/L}$

8.8 Stress-Strain Curve

A typical stress-strain curve for a ductile material (like metal wire) shows:

1. OA — Proportional limit: Linear region, Hooke's law obeyed
2. A — Elastic limit: Beyond this, permanent deformation begins
3. BC — Plastic region: Large strain for small increase in stress
4. D — Ultimate tensile strength: Maximum stress the material can withstand
5. E — Fracture/Breaking point: Material breaks

Breaking stress: The stress at the fracture point where the material actually breaks. Also called fracture stress. Beyond this point, the material loses structural integrity completely and cannot be restored.

8.9 Elastic After-Effect and Fatigue

Elastic after-effect: The delay in a material returning to its original shape after removal of the deforming force.

Elastic fatigue: Weakening of a material due to repeated cycles of loading and unloading.

INTEXT QUESTIONS 8.1

Q1. What will be the nature of inter-atomic forces when deforming force applied on an object (i) increases, (ii) decreases the inter-atomic separation?

Ans:

(i) When deforming force increases inter-atomic separation ($R>R_0$): The inter-atomic forces become attractive in nature. These attractive forces try to bring the atoms back to their equilibrium positions. The atoms resist the increase in separation.

(ii) When deforming force decreases inter-atomic separation ($R<R_0$): The inter-atomic forces become repulsive in nature. These repulsive forces try to push the atoms back to their equilibrium positions. The atoms resist being compressed further.

Q2. If we clamp a rod rigidly at one end and a force is applied normally to its cross section at the other end, name the type of stress and strain.

Ans:

• Type of Stress: Longitudinal stress (tensile if pulling, compressive if pushing). The stress acts along the length of the rod. Stress = $F/A$, where $F$ is the applied force and $A$ is the cross-sectional area.
• Type of Strain: Linear strain (longitudinal strain). Strain = $\Delta L/L$, where $\Delta L$ is change in length and $L$ is original length.

Q3. The ratio of stress to strain remains constant for small deformation of a metal wire. For large deformations what will be the changes in this ratio?

Ans: For small deformations, the ratio of stress to strain remains constant according to Hooke's Law (modulus of elasticity).

For large deformations:

• The ratio of stress to strain decreases
• The wire crosses its elastic limit
• Beyond the elastic limit, the material enters the plastic region
• In the plastic region, stress increases more slowly compared to strain
• The stress-strain relationship becomes non-linear
• Eventually, the wire may reach the breaking point where it fractures

Q4. Under what conditions, a stress is known as breaking stress?

Ans: A stress is known as breaking stress under these conditions:

• At the fracture point: When applied stress reaches the maximum value the material can withstand before breaking
• Beyond ultimate tensile strength: The stress at which the material actually breaks
• Point of fracture on stress-strain curve: Corresponds to the breaking point
• Material failure: The material loses its structural integrity completely
• No recovery possible: The material cannot be restored to its original form

Breaking stress is also called fracture stress or ultimate breaking strength.

Q5. If mass of 4 kg is attached to the end of a vertical wire of length 4 m with a diameter 0.64 mm, the extension is 0.60 mm. Calculate the tensile stress and strain.

Ans:

• $m=4$ kg, $L=4$ m, $d=0.64$ mm = $0.64\times 10^{-3}$ m
• $\Delta L=0.60$ mm = $0.60\times 10^{-3}$ m, $g=9.8$ m/s²

Tensile Stress: $F=mg=4\times 9.8=39.2\text{ N}$ $A=\frac{\pi d^2}{4}=\frac{\pi \times (0.64\times 10^{-3}{)}^2}{4}=3.22\times 10^{-7}{\text{ m}}^2$ $\text{Stress}=\frac{F}{A}=\frac{39.2}{3.22\times 10^{-7}}=1.22\times 10^8{\text{ N/m}}^2$

Tensile Strain: $\text{Strain}=\frac{\Delta L}{L}=\frac{0.60\times 10^{-3}}{4}=1.5\times 10^{-4}$

Answer: Tensile stress = $1.22\times 10^8$ N/m², Tensile strain = $1.5\times 10^{-4}$

Terminal Exercise

1. Distinguish between elasticity and plasticity. Give two examples of elastic and plastic materials.

2. Define stress and strain. What are the different types of each?

3. State Hooke's law. What is the elastic limit? Draw a labelled stress-strain curve for a ductile material.

4. Define Young's modulus. Derive the expression $Y=\frac{FL}{A\Delta L}$.

5. A steel wire of length 4.7 m and cross-sectional area $3.0\times 10^{-5}$ m² stretches by the same amount as a copper wire of length 3.5 m and cross-sectional area $4.0\times 10^{-5}$ m² under a given load. What is the ratio of Young's modulus of steel to that of copper?

6. Define bulk modulus and compressibility. Show that for an ideal gas at constant temperature, bulk modulus equals the pressure.

7. Define shear modulus. A cube of aluminium of side 10 cm is subjected to a shearing force of 100 N. The top surface is displaced by 0.02 cm relative to the bottom. Calculate the shear modulus.

8. Define Poisson's ratio. Why is it always less than 0.5?

9. A mass of 10 kg is attached to the free end of a brass wire of length 2 m and diameter 1 mm. The wire stretches by 2 mm. Calculate: (a) stress, (b) strain, (c) Young's modulus of brass. (g = 10 m/s²)

10. Explain the terms: (a) elastic after-effect, (b) elastic fatigue, (c) breaking stress.

11. Why are bridges declared unsafe after long use even though they may not show visible damage?

12. Two wires A and B are of the same material. Their lengths are in the ratio 1:2 and diameters are in the ratio 2:1. If they are stretched by the same force, find the ratio of their extensions.

Worked Examples

Example 1: Young's Modulus

Problem: A wire of length 2 m and cross-sectional area $10^{-6}$ m² is stretched by 1 mm under a load of 10 kg. Find Young's modulus. (g = 10 m/s²)

Solution: $Y=\frac{FL}{A\Delta L}=\frac{(10\times 10)\times 2}{10^{-6}\times 10^{-3}}=\frac{200}{10^{-9}}=2\times 10^{11}{\text{ N/m}}^2$

Example 2: Stress and Strain

Problem: A copper wire of length 3 m and radius 0.5 mm is stretched by a load of 5 kg. Find the stress and strain. ($Y_{\text{Cu}}=1.1\times 10^{11}$ N/m², g = 10 m/s²)

Solution: $F=5\times 10=50\text{ N}$ $A=\pi r^2=\pi \times (0.5\times 10^{-3}{)}^2=7.85\times 10^{-7}{\text{ m}}^2$ $\text{Stress}=\frac{50}{7.85\times 10^{-7}}=6.37\times 10^7{\text{ N/m}}^2$ $\text{Strain}=\frac{\text{Stress}}{Y}=\frac{6.37\times 10^7}{1.1\times 10^{11}}=5.79\times 10^{-4}$

Example 3: Bulk Modulus

Problem: A solid sphere of volume 0.5 m³ is placed in water at a depth where the pressure is $2\times 10^5$ Pa greater than at the surface. If the volume decreases by 0.01%, find the bulk modulus.

Solution: $\frac{\Delta V}{V}=\frac{0.01}{100}=10^{-4}$ $B=-\frac{\Delta P}{\Delta V/V}=\frac{2\times 10^5}{10^{-4}}=2\times 10^9{\text{ N/m}}^2$

Common Mistakes

1. Confusing stress with pressure: Stress is restoring force per unit area within a material; pressure is external force per unit area.
2. Using Hooke's law beyond the elastic limit: It's valid only for small deformations.
3. Forgetting the negative sign in bulk modulus: It ensures B is positive since $\Delta V$ is negative when $\Delta P$ is positive.
4. Confusing breaking stress with ultimate tensile strength: Ultimate tensile strength is the maximum stress before necking; breaking stress is at fracture.
5. Mixing up the three moduli: Young's (linear), Bulk (volumetric), Shear (tangential) — different types of deformation.

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