Non-Rationalised Science NCERT Notes and Solutions (Class 6th to 10th)
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Non-Rationalised Science NCERT Notes and Solutions (Class 11th)
Physics		Chemistry			Biology
Non-Rationalised Science NCERT Notes and Solutions (Class 12th)
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Class 11th (Physics) Chapters
1. Physical World	2. Units And Measurements	3. Motion In A Straight Line
4. Motion In A Plane	5. Laws Of Motion	6. Work, Energy And Power
7. System Of Particles And Rotational Motion	8. Gravitation	9. Mechanical Properties Of Solids
10. Mechanical Properties Of Fluids	11. Thermal Properties Of Matter	12. Thermodynamics
13. Kinetic Theory	14. Oscillations	15. Waves

Chapter 9 Mechanical Properties Of Solids

Introduction

While we often treat objects as ideal rigid bodies with a definite, unchanging shape and size, real-world solid objects can be deformed. They can be stretched, compressed, or bent when a sufficiently large external force is applied. This chapter explores the mechanical properties of solids, focusing on how they deform under applied forces.

The property of a body by virtue of which it tends to regain its original size and shape after a deforming force is removed is known as elasticity. The deformation that disappears upon removal of the force is called an elastic deformation. Materials like steel and rubber exhibit elastic behaviour.

In contrast, substances like putty or mud, which do not show a tendency to regain their original shape and get permanently deformed, are called plastic. This property is known as plasticity.

Understanding the elastic behaviour of materials is crucial in engineering and design, from constructing buildings and bridges to designing automobiles and artificial limbs. This chapter will provide the foundational concepts to answer questions about the strength, shape, and behaviour of materials under load.

Elastic Behaviour of Solids

The elastic behaviour of solids can be understood at a microscopic level. In a solid, atoms or molecules are held together in stable equilibrium positions by interatomic or intermolecular forces. These forces can be visualized as springs connecting the atoms.

A model of a solid where atoms are represented by balls and the interatomic forces are represented by springs connecting them.

When an external deforming force is applied, the atoms are displaced from their equilibrium positions, causing the "springs" (interatomic bonds) to be stretched or compressed. This displacement develops an internal restoring force within the solid. When the external deforming force is removed, this internal restoring force drives the atoms back to their original equilibrium positions, and the body regains its original shape and size. This microscopic mechanism is the origin of elasticity.

English physicist Robert Hooke was a pioneer in studying this behaviour, leading to his famous law of elasticity.

Stress and Strain

To quantify the elastic properties of materials, we define two key concepts: stress and strain.

Stress

When a body is subjected to a deforming force, internal restoring forces are developed within it. In equilibrium, the restoring force is equal in magnitude and opposite in direction to the applied deforming force.

Stress is defined as the internal restoring force per unit area. If a force F is applied to a cross-sectional area A:

$\text{Stress} (\sigma) = \frac{\text{Force}}{\text{Area}} = \frac{F}{A}$

The SI unit of stress is N/m² or Pascal (Pa).
Its dimensional formula is $[ML^{-1}T^{-2}]$.

Strain

Strain is the measure of the deformation produced in a body. It is defined as the fractional change in a dimension (length, volume, or shape).

$\text{Strain} (\epsilon) = \frac{\text{Change in Dimension}}{\text{Original Dimension}}$

Since strain is a ratio of similar quantities, it is a dimensionless and unitless quantity.

Types of Stress and Strain

1. Longitudinal Stress and Strain

This occurs when forces are applied normal (perpendicular) to the cross-sectional area, causing a change in length.

Tensile Stress: If the body is stretched.
Compressive Stress: If the body is compressed.
Longitudinal Strain: The change in length ($\Delta L$) divided by the original length ($L$).

$\text{Longitudinal Strain} = \frac{\Delta L}{L}$

A cylinder being stretched by equal and opposite forces, resulting in an elongation ΔL. This is an example of tensile stress and longitudinal strain.

2. Shearing Stress and Strain

This occurs when forces are applied parallel (tangential) to the cross-sectional area, causing a change in shape.

Shearing Stress: The tangential force per unit area.
Shearing Strain: The ratio of the relative displacement of the faces ($\Delta x$) to the length of the body ($L$). It is also given by the angle of shear, $\theta$.

$\text{Shearing Strain} = \frac{\Delta x}{L} = \tan\theta \approx \theta \quad \text{(for small angles)}$

A cylinder subjected to tangential forces, causing its top face to displace by Δx relative to the bottom face, resulting in a shear angle θ.

class="blackheading">3. Hydraulic Stress and Volume Strain

This occurs when a body is subjected to a uniform pressure from all sides, such as when submerged in a fluid.

Hydraulic Stress: The restoring force per unit area, which is equal in magnitude to the applied pressure ($p$).
Volume Strain: The change in volume ($\Delta V$) divided by the original volume ($V$).

$\text{Volume Strain} = \frac{\Delta V}{V}$

A sphere submerged in a fluid under high pressure, causing its volume to decrease by ΔV without a change in shape.

Hooke’s Law

For most materials, when the deformations are small, there is a simple linear relationship between stress and strain. This empirical relationship is known as Hooke's Law.

Hooke's Law: Within the elastic limit, the stress developed in a body is directly proportional to the strain produced in it.

Mathematically:

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

$\text{Stress} = k \times \text{Strain}$

The constant of proportionality, $k$, is called the modulus of elasticity. Its value is a characteristic property of the material and indicates its stiffness. A higher modulus of elasticity means a stiffer material (more stress is required to produce a given strain).

Stress-Strain Curve

The relationship between stress and strain for a material can be determined experimentally by stretching a wire of the material and measuring the deformation. A plot of stress versus strain gives a characteristic curve for the material.

$A typical stress-strain curve for a ductile metal, showing the elastic region, yield point, plastic region, ultimate tensile strength, and fracture point.$

Key features of a typical stress-strain curve for a metal:

Proportional Limit (Region O to A): The curve is a straight line, and Hooke's law is obeyed. The material is elastic.
Elastic Limit or Yield Point (Point B): This is the point up to which the material behaves elastically. If the load is removed before this point, the body returns to its original dimensions. The stress at this point is the yield strength ($\sigma_y$).
Plastic Region (Region B to D): If the material is loaded beyond the elastic limit, it undergoes permanent deformation. Even after the load is removed, the material has a permanent set (it does not return to its original shape). This is called plastic deformation.
Ultimate Tensile Strength (Point D): This is the maximum stress that the material can withstand before it starts to fail.
Fracture Point (Point E): The point at which the material breaks.

Ductile vs. Brittle Materials

Ductile materials (like copper, mild steel) have a large plastic region. The ultimate strength and fracture points are far apart. They can be drawn into wires.
Brittle materials (like glass, cast iron) have a very small or non-existent plastic region. The ultimate strength and fracture points are very close. They fracture soon after the elastic limit is reached.

Elastomers

Substances like rubber or the elastic tissue of the aorta can be stretched to very large strains and still return to their original shape. These materials are called elastomers. Their stress-strain curve is not linear and they do not obey Hooke's law, but they do have a large elastic region.

Elastic Moduli

The modulus of elasticity is the ratio of stress to strain within the elastic limit. There are three types, corresponding to the three types of stress and strain.

1. Young’s Modulus (Y)

This relates to longitudinal stress and strain. It is a measure of a material's resistance to a change in its length.

Young's Modulus (Y) = $\frac{\text{Longitudinal Stress}}{\text{Longitudinal Strain}}$

$Y = \frac{\sigma}{\epsilon} = \frac{F/A}{\Delta L/L} = \frac{F L}{A \Delta L}$

The SI unit of Young's modulus is N/m² or Pa. Steel has a very high Young's modulus, making it very stiff and suitable for structural applications.

Example 1. A structural steel rod has a radius of 10 mm and a length of 1.0 m. A 100 kN force stretches it along its length. Calculate (a) stress, (b) elongation, and (c) strain on the rod. Young’s modulus of structural steel is $2.0 \times 10^{11} \text{ N m}^{-2}$.

Answer:

Given: $r = 10 \text{ mm} = 10^{-2} \text{ m}$, $L = 1.0 \text{ m}$, $F = 100 \text{ kN} = 10^5 \text{ N}$, $Y = 2.0 \times 10^{11} \text{ Pa}$.

Area of cross-section, $A = \pi r^2 = \pi (10^{-2})^2 = \pi \times 10^{-4} \text{ m}^2$.

(a) Stress:

Stress = $\frac{F}{A} = \frac{10^5 \text{ N}}{\pi \times 10^{-4} \text{ m}^2} \approx 3.18 \times 10^8 \text{ Pa}$.

(b) Elongation ($\Delta L$):

From $Y = \frac{F L}{A \Delta L}$, we get $\Delta L = \frac{F L}{A Y} = \frac{\text{Stress} \times L}{Y}$.

$\Delta L = \frac{(3.18 \times 10^8 \text{ Pa}) \times (1.0 \text{ m})}{2.0 \times 10^{11} \text{ Pa}} = 1.59 \times 10^{-3} \text{ m} = 1.59 \text{ mm}$.

(c) Strain:

Strain = $\frac{\Delta L}{L} = \frac{1.59 \times 10^{-3} \text{ m}}{1.0 \text{ m}} = 1.59 \times 10^{-3}$. (This is equivalent to 0.159%).

2. Shear Modulus (G) or Modulus of Rigidity

This relates to shearing stress and strain. It is a measure of a material's resistance to a change in its shape.

Shear Modulus (G) = $\frac{\text{Shearing Stress}}{\text{Shearing Strain}}$

$G = \frac{\sigma_s}{\epsilon_s} = \frac{F/A}{\Delta x/L} = \frac{F/A}{\theta}$

For most materials, the shear modulus is approximately one-third of the Young's modulus ($G \approx Y/3$).

3. Bulk Modulus (B)

This relates to hydraulic stress and volume strain. It is a measure of a material's resistance to a change in its volume (incompressibility).

Bulk Modulus (B) = $\frac{\text{Hydraulic Stress}}{\text{Volume Strain}}$

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

The negative sign indicates that an increase in pressure (positive $p$) leads to a decrease in volume (negative $\Delta V$), ensuring that B is always positive.

The reciprocal of the bulk modulus, $k=1/B$, is called compressibility. Solids are the least compressible (high B), while gases are the most compressible (low B).

Poisson's Ratio ($\sigma$)

When a wire is stretched along its length (longitudinal strain), it gets slightly thinner in its cross-section (lateral strain). Poisson's ratio is the ratio of the lateral strain to the longitudinal strain within the elastic limit.

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

It is a dimensionless quantity. For most steels, its value is between 0.28 and 0.30.

Applications of Elastic Behaviour of Materials

The knowledge of elastic properties is vital in many engineering applications to ensure structural safety and efficiency.

1. Design of Cranes and Ropes

When designing a crane rope to lift a heavy load, it must be ensured that the stress in the rope does not exceed the elastic limit of the material, to avoid permanent deformation. A safety factor is typically included.

For a load of 10 metric tons ($10^4$ kg), the stress in a steel rope must be less than the yield strength of steel ($\approx 300 \times 10^6 \text{ Pa}$).

Minimum Area, $A \ge \frac{\text{Force}}{\text{Yield Strength}} = \frac{Mg}{\sigma_y} = \frac{10^4 \text{ kg} \times 9.8 \text{ m/s}^2}{300 \times 10^6 \text{ N/m}^2} \approx 3.3 \times 10^{-4} \text{ m}^2$.

This corresponds to a radius of about 1 cm. To provide flexibility and strength, such ropes are made by braiding many thin wires together.

2. Design of Bridges and Beams

Beams used in bridges and buildings are designed to minimize bending (or sagging) under a load. The sag, $\delta$, of a beam of length $l$, breadth $b$, and depth $d$, supported at its ends and loaded at the center with a weight $W$ is given by:

$\delta = \frac{Wl^3}{4Ybd^3}$

To minimize bending, one should:

Use a material with a high Young's modulus (Y).
Make the beam's depth (d) large, as the sag is inversely proportional to $d^3$.

This is why beams used in construction often have an I-shaped cross-section. This shape provides a large depth and a wide load-bearing surface, making it strong against bending and buckling while being lighter and more cost-effective than a solid rectangular beam of the same depth.

An I-shaped cross-section for a beam, which is a common design to maximize strength against bending while minimizing weight.

3. Height of Mountains

The elastic properties of rocks limit the maximum height of mountains on Earth. The pressure at the base of a mountain of height $h$ and density $\rho$ is approximately $h\rho g$. This pressure creates a shearing stress within the rock. If this stress exceeds the elastic limit of the rock (around $30 \times 10^7$ Pa), the rock will begin to flow like a plastic material. This calculation yields a maximum possible height of about 10 km, which is close to the height of Mount Everest.

Exercises

Question 9.1. A steel wire of length 4.7 m and cross-sectional area $3.0 \times 10^{-5} \text{ m}^2$ stretches by the same amount as a copper wire of length 3.5 m and cross-sectional area of $4.0 \times 10^{–5} \text{ m}^2$ under a given load. What is the ratio of the Young’s modulus of steel to that of copper?

Answer:

Question 9.2. Figure 9.11 shows the strain-stress curve for a given material. What are (a) Young’s modulus and (b) approximate yield strength for this material?

A stress-strain curve. Stress (in 10^6 N m^-2) is on the y-axis and strain is on the x-axis. The curve is a straight line from the origin up to a stress of about 150, at which point the strain is 0.002. Then the curve bends over.

Answer:

Question 9.3. The stress-strain graphs for materials A and B are shown in Fig. 9.12.

Two stress-strain graphs for materials A and B, drawn on the same axes. Both are straight lines starting from the origin. The line for material A has a steeper slope than the line for material B.

The graphs are drawn to the same scale.

(a) Which of the materials has the greater Young’s modulus?

(b) Which of the two is the stronger material?

Answer:

Question 9.4. Read the following two statements below carefully and state, with reasons, if it is true or false.

(a) The Young’s modulus of rubber is greater than that of steel;

(b) The stretching of a coil is determined by its shear modulus.

Answer:

Question 9.5. Two wires of diameter 0.25 cm, one made of steel and the other made of brass are loaded as shown in Fig. 9.13. The unloaded length of steel wire is 1.5 m and that of brass wire is 1.0 m. Compute the elongations of the steel and the brass wires.

A setup showing two wires, steel and brass, suspended from a rigid support. A 4 kg mass hangs from the steel wire. The brass wire hangs from the bottom of the 4 kg mass, and a 6 kg mass hangs from the brass wire.

Answer:

Question 9.6. The edge of an aluminium cube is 10 cm long. One face of the cube is firmly fixed to a vertical wall. A mass of 100 kg is then attached to the opposite face of the cube. The shear modulus of aluminium is 25 GPa. What is the vertical deflection of this face?

Answer:

Question 9.7. Four identical hollow cylindrical columns of mild steel support a big structure of mass 50,000 kg. The inner and outer radii of each column are 30 and 60 cm respectively. Assuming the load distribution to be uniform, calculate the compressional strain of each column.

Answer:

Question 9.8. A piece of copper having a rectangular cross-section of 15.2 mm × 19.1 mm is pulled in tension with 44,500 N force, producing only elastic deformation. Calculate the resulting strain?

Answer:

Question 9.9. A steel cable with a radius of 1.5 cm supports a chairlift at a ski area. If the maximum stress is not to exceed $10^8 \text{ N m}^{–2}$, what is the maximum load the cable can support ?

Answer:

Question 9.10. A rigid bar of mass 15 kg is supported symmetrically by three wires each 2.0 m long. Those at each end are of copper and the middle one is of iron. Determine the ratios of their diameters if each is to have the same tension.

Answer:

Question 9.11. A 14.5 kg mass, fastened to the end of a steel wire of unstretched length 1.0 m, is whirled in a vertical circle with an angular velocity of 2 rev/s at the bottom of the circle. The cross-sectional area of the wire is 0.065 cm$^2$. Calculate the elongation of the wire when the mass is at the lowest point of its path.

Answer:

Question 9.12. Compute the bulk modulus of water from the following data: Initial volume = 100.0 litre, Pressure increase = 100.0 atm (1 atm = $1.013 \times 10^5$ Pa), Final volume = 100.5 litre. Compare the bulk modulus of water with that of air (at constant temperature). Explain in simple terms why the ratio is so large.

Answer:

Question 9.13. What is the density of water at a depth where pressure is 80.0 atm, given that its density at the surface is $1.03 \times 10^3 \text{ kg m}^{–3}$?

Answer:

Question 9.14. Compute the fractional change in volume of a glass slab, when subjected to a hydraulic pressure of 10 atm.

Answer:

Question 9.15. Determine the volume contraction of a solid copper cube, 10 cm on an edge, when subjected to a hydraulic pressure of $7.0 \times 10^6$ Pa.

Answer:

Question 9.16. How much should the pressure on a litre of water be changed to compress it by 0.10%?

Answer:

Additional Exercises

Question 9.17. Anvils made of single crystals of diamond, with the shape as shown in Fig. 9.14, are used to investigate behaviour of materials under very high pressures. Flat faces at the narrow end of the anvil have a diameter of 0.50 mm, and the wide ends are subjected to a compressional force of 50,000 N. What is the pressure at the tip of the anvil?

A diamond anvil, which is a truncated cone shape, used for creating high pressure.

Answer:

Question 9.18. A rod of length 1.05 m having negligible mass is supported at its ends by two wires of steel (wire A) and aluminium (wire B) of equal lengths as shown in Fig. 9.15. The cross-sectional areas of wires A and B are $1.0 \text{ mm}^2$ and $2.0 \text{ mm}^2$, respectively. At what point along the rod should a mass m be suspended in order to produce (a) equal stresses and (b) equal strains in both steel and aluminium wires.

A horizontal rod is supported by a steel wire (A) at one end and an aluminum wire (B) at the other end. A mass m is suspended from the rod.

Answer:

Question 9.19. A mild steel wire of length 1.0 m and cross-sectional area $0.50 \times 10^{-2} \text{ cm}^2$ is stretched, well within its elastic limit, horizontally between two pillars. A mass of 100 g is suspended from the mid-point of the wire. Calculate the depression at the midpoint.

Answer:

Question 9.20. Two strips of metal are riveted together at their ends by four rivets, each of diameter 6.0 mm. What is the maximum tension that can be exerted by the riveted strip if the shearing stress on the rivet is not to exceed $6.9 \times 10^7$ Pa? Assume that each rivet is to carry one quarter of the load.

Answer:

Question 9.21. The Marina trench is located in the Pacific Ocean, and at one place it is nearly eleven km beneath the surface of water. The water pressure at the bottom of the trench is about $1.1 \times 10^8$ Pa. A steel ball of initial volume 0.32 m$^3$ is dropped into the ocean and falls to the bottom of the trench. What is the change in the volume of the ball when it reaches to the bottom?

Answer: