Physics: Class 11 (Book 2) — Chapter 8: Mechanical Properties of Solids

• elasticity : The property of a body by virtue of which it tends to regain its original size and shape when the applied deforming force is removed.
• elastic deformation : The temporary change in shape or size of a body that disappears completely once the deforming forces are removed.
• plasticity : The property of a body by which it gets permanently deformed and shows no tendency to regain its original shape and size after the deforming force is removed.
• plastic deformation : The permanent change in shape or size of a body. Materials like putty and mud are close to ideal plastics.

The restoring force per unit area is defined as stress .

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

• SI Unit: N m⁻² or pascal (Pa)
• Dimensional Formula: $[ML^{-1}T^{-2}]$

1. Longitudinal Stress and Strain:

   When deforming forces act normal to the cross-sectional area of a cylinder, changing its length:

   • tensile stress : Restoring force per unit area when the cylinder is stretched.
   • compressive stress : Restoring force per unit area when the cylinder is compressed.
   • longitudinal strain : The ratio of change in length ($\Delta L$) to the original length ($L$).

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

2. Shearing (Tangential) Stress and Strain:

   When equal and opposite deforming forces are applied parallel to the cross-sectional area, causing relative displacement between opposite faces:

   • shearing stress : Restoring force per unit area developed due to applied tangential force.
   • shearing strain : The ratio of relative displacement of the faces ($\Delta x$) to the length ($L$).

   $$\text{Shearing strain} = \frac{\Delta x}{L} = \tan \theta \approx \theta$$

   where $\theta$ is the angular displacement of the cylinder from its original vertical position.

3. Hydraulic Stress and Volume Strain:

   When a solid body is submerged in a fluid, it experiences uniform high pressure on all sides, leading to a decrease in volume without changing its geometrical shape:

   • hydraulic stress : The internal restoring force per unit area, equal in magnitude to the hydraulic pressure ($p$).
   • volume strain : The ratio of change in volume ($\Delta V$) to the original volume ($V$).

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

   Since strain is a ratio of similar physical quantities, it is dimensionless and has no units.

For small deformations, stress and strain are directly proportional to each other. This empirical relation is known as Hooke's law :

$$\text{stress} \propto \text{strain}$$

$$\text{stress} = k \times \text{strain}$$

where $k$ is the constant of proportionality, known as the modulus of elasticity .

A typical stress-strain curve for a metal under tensile load exhibits several distinct regions:

• Region O to A (Proportional Limit): The curve is linear. Hooke's law is strictly obeyed. The body behaves as a perfectly elastic body and regains its original dimensions when the load is removed.
• Region A to B (Yield Point / Elastic Limit): Stress and strain are not proportional, but the body still returns to its original dimensions when the load is removed. Point B is the yield point (or elastic limit ), and the corresponding stress is the yield strength ($\sigma_y$).
• Region B to D (Plastic Deformation): If the load is increased further, stress exceeds the yield strength, and strain increases rapidly. If the load is removed at a point C between B and D, the material does not regain its original length and retains a permanent set (residual strain).
• Point D (Ultimate Tensile Strength): This represents the maximum stress the material can withstand, denoted as ultimate tensile strength ($\sigma_u$).
• Point E (Fracture Point): Beyond point D, additional strain is produced even by a reduced force, and the material breaks at the fracture point E.
• Brittle vs. Ductile: If points D and E are close, the material is brittle (e.g., glass). If they are far apart, the material is ductile (e.g., copper, steel).

Substances like rubber or the elastic tissue of the aorta can be stretched to undergo very large strains but do not obey Hooke's law even for small deformations. They do not have a well-defined plastic region and are called elastomers .

The ratio of tensile (or compressive) stress ($\sigma$) to longitudinal strain ($\epsilon$) is defined as Young's modulus :

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

Since strain is dimensionless, the unit of Young's modulus is the same as that of stress: N m⁻² or Pa .

Mathematical Problem Statement: 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 $2.0 \times 10^{11}\text{ N m}^{-2}$)

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Step-by-Step Algorithmic Solution Trace:

1. Calculate the cross-sectional area of the rod: $$A = \pi r^2 = 3.14 \times (10 \times 10^{-3}\text{ m})^2 = 3.14 \times 10^{-4}\text{ m}^2$$
2. Calculate the stress: $$\text{Stress} = \frac{F}{A} = \frac{100 \times 10^3\text{ N}}{3.14 \times 10^{-4}\text{ m}^2} = 3.18 \times 10^8\text{ N m}^{-2}$$
3. Calculate the elongation ($\Delta L$): $$\Delta L = \frac{(F/A)L}{Y} = \frac{(3.18 \times 10^8\text{ N m}^{-2})(1.0\text{ m})}{2.0 \times 10^{11}\text{ N m}^{-2}} = 1.59 \times 10^{-3}\text{ m} = 1.59\text{ mm}$$
4. Calculate the strain: $$\text{Strain} = \frac{\Delta L}{L} = \frac{1.59 \times 10^{-3}\text{ m}}{1.0\text{ m}} = 1.59 \times 10^{-3} = 0.16\%$$

Final Evaluation Output: Stress = $3.18 \times 10^8\text{ N m}^{-2}$, Elongation = 1.59 mm , Strain = 0.16%

The ratio of shearing stress to shearing strain is called the shear modulus (or modulus of rigidity ), represented by $G$:

$$G = \frac{\text{shearing stress (}\sigma_s\text{)}}{\text{shearing strain}} = \frac{F/A}{\Delta x/L} = \frac{F \times L}{A \times \Delta x}$$

Alternatively, using angular deformation $\theta$:

$$G = \frac{F/A}{\theta} = \frac{F}{A \times \theta}$$

For most materials, the shear modulus is related to Young's modulus by $G \approx Y/3$.

The ratio of hydraulic stress to the corresponding volume strain is called the bulk modulus , represented by $B$:

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

The negative sign indicates that an increase in pressure results in a decrease in volume. For a system in equilibrium, $B$ is always positive.

The reciprocal of the bulk modulus is called compressibility ($k$):

$$k = \frac{1}{B} = -\frac{1}{\Delta p} \left(\frac{\Delta V}{V}\right)$$

Solids have very high bulk moduli (least compressible), liquids have intermediate values, and gases have very low bulk moduli (highly compressible).

Mathematical Problem Statement: The average depth of the Indian Ocean is about 3000 m . Calculate the fractional compression, $\Delta V/V$, of water at the bottom of the ocean, given that the bulk modulus of water is $2.2 \times 10^9\text{ N m}^{-2}$ and $g = 10\text{ m s}^{-2}$.

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Step-by-Step Algorithmic Solution Trace:

1. Calculate the hydrostatic pressure ($p$) at depth $h = 3000\text{ m}$: $$p = h \rho g = 3000\text{ m} \times 1000\text{ kg m}^{-3} \times 10\text{ m s}^{-2} = 3 \times 10^7\text{ N m}^{-2}$$
2. Calculate the fractional compression ($\Delta V/V$): $$\frac{\Delta V}{V} = \frac{p}{B} = \frac{3 \times 10^7\text{ N m}^{-2}}{2.2 \times 10^9\text{ N m}^{-2}} = 1.36 \times 10^{-2} = 1.36\%$$

Final Evaluation Output: Fractional compression $\Delta V/V = 1.36\%$

When a wire is stretched, its length increases while its diameter contracts. The strain perpendicular to the applied force is called lateral strain.

The ratio of lateral strain to longitudinal strain is called Poisson's ratio ($\sigma$):

$$\sigma = \frac{\text{Lateral strain}}{\text{Longitudinal strain}} = \frac{\Delta d/d}{\Delta L/L} = \frac{\Delta d}{\Delta L} \times \frac{L}{d}$$

Poisson's ratio is a dimensionless pure number. For steel, its value lies between 0.28 and 0.30 , and for aluminium alloys, it is about 0.33 .

Work done against inter-atomic forces during tensile deformation is stored in the wire as elastic potential energy ($W$):

$$W = \frac{1}{2} \times \text{stress} \times \text{strain} \times \text{volume of the wire}$$

The elastic potential energy per unit volume ($u$) is given by:

$$u = \frac{1}{2} \sigma \epsilon$$

where $\sigma$ is stress and $\epsilon$ is strain.

To lift a load of mass $M = 10\text{ tonnes} = 10^4\text{ kg}$ without exceeding the elastic limit of steel (yield strength $\sigma_y \approx 300 \times 10^6\text{ N m}^{-2}$):

$$A \ge \frac{W}{\sigma_y} = \frac{Mg}{\sigma_y} = \frac{10^4\text{ kg} \times 9.8\text{ m s}^{-2}}{300 \times 10^6\text{ N m}^{-2}} = 3.3 \times 10^{-4}\text{ m}^2$$

This corresponds to a single solid wire of radius $r \approx 1\text{ cm}$. To ensure a safety factor of 10, a thicker rope of radius 3 cm is recommended. Ropes are made of multiple braided thin wires to ensure flexibility and strength.

A beam of length $l$, breadth $b$, and depth $d$, supported at its ends and loaded at the centre with a load $W$, sags by an amount $\delta$:

$$\delta = \frac{W l^3}{4 b d^3 Y}$$

To minimize sagging ($\delta$):

• Use a material with a large Young's modulus ($Y$).
• Increasing depth ($d$) is highly effective since $\delta \propto d^{-3}$.
• However, deep narrow beams can bend sideways under load, a failure mode known as buckling .
• To prevent buckling while minimizing weight and cost, beams are designed with an I-shaped cross-section .

The maximum height of a mountain on Earth is limited by the elastic properties of rocks at its base. The shear stress at the bottom of a mountain of height $h$ and density $\rho$ is approximately $h \rho g$.

For typical rock, the elastic limit is $30 \times 10^7\text{ N m}^{-2}$ and density $\rho \approx 3 \times 10^3\text{ kg m}^{-3}$:

$$h \rho g \le 30 \times 10^7\text{ N m}^{-2}$$

$$h \le \frac{30 \times 10^7\text{ N m}^{-2}}{3 \times 10^3\text{ kg m}^{-3} \times 10\text{ m s}^{-2}} = 10\text{ km}$$

This explains why the height of Mt. Everest (~8.8 km) is close to the theoretical limit of 10 km .