Mechanical Properties of Solids: CBSE Class 11 Physics NCERT Guide

Understanding Stress, Strain, and Elasticity

Solids, unlike liquids and gases, possess a definite shape and volume. However, when an external force, known as a deforming force, is applied to a solid body, its configuration (shape or size) changes. The ability of a body to regain its original configuration after the removal of the deforming force is called elasticity. Materials that exhibit this property are known as elastic bodies. Conversely, if a body does not regain its original configuration and undergoes a permanent deformation, it is said to be a plastic body, and the phenomenon is called plasticity.

To understand this behavior quantitatively, we define two key terms: Stress and Strain.

1. Stress (σ): When a deforming force is applied, internal restoring forces are set up within the body, opposing the deformation. Stress is defined as the restoring force developed per unit cross-sectional area of the body. Mathematically, it's given by:

Stress = Restoring Force / Area (σ = F/A)
Its SI unit is N/m² or Pascal (Pa). Stress can be normal (perpendicular to the surface, causing change in length or volume, like tensile or compressive stress) or tangential/shear (parallel to the surface, causing change in shape).

1. Strain (ε): Strain is a measure of the deformation produced in the body relative to its original dimensions. It is defined as the ratio of the change in configuration to the original configuration. Since it is a ratio of two similar quantities, strain is a dimensionless quantity and has no units.

• Longitudinal Strain: ΔL / L (change in length / original length).
• Volumetric Strain: ΔV / V (change in volume / original volume).
• Shear Strain: Δx / L = tan θ ≈ θ (relative displacement of faces / original length, where θ is the angle of shear).

Hooke's Law: Within the elastic limit, stress is directly proportional to strain. This is a fundamental law in elasticity.
Stress ∝ Strain or Stress = E × Strain
Here, E is the constant of proportionality, known as the modulus of elasticity, which depends on the material and the type of deformation.

Key Moduli of Elasticity

Young's Modulus (Y)

It is the ratio of normal stress to longitudinal strain. It quantifies the resistance of a material to elastic deformation under tension or compression. Y = (Normal Stress) / (Longitudinal Strain) = (F/A) / (ΔL/L).

Bulk Modulus (K)

It is the ratio of normal stress (pressure) to volumetric strain. It measures a substance's resistance to compression and changes in volume. K = (Normal Stress) / (Volumetric Strain) = P / (ΔV/V). The reciprocal of bulk modulus is compressibility.

Shear Modulus (G) or Modulus of Rigidity

It is the ratio of tangential stress to shear strain. It describes the material's resistance to shearing deformation, i.e., deformation in which layers slide parallel to each other. G = (Tangential Stress) / (Shear Strain) = (F_tangential/A) / θ.

Poisson's Ratio (ν)

When a wire is stretched longitudinally, its length increases, but its diameter decreases. Poisson's ratio is defined as the ratio of lateral strain to longitudinal strain. ν = - (Lateral Strain) / (Longitudinal Strain) = - (ΔD/D) / (ΔL/L). The negative sign indicates that lateral contraction accompanies longitudinal extension.

Fully Worked Examples

• Example 1: Calculating Elongation of a Wire A steel wire of length 2.0 m and cross-sectional area 1.0 × 10⁻⁶ m² is stretched by a force of 100 N. If Young's modulus for steel is 2.0 × 10¹¹ N/m², calculate the elongation (increase in length) of the wire. Step 1: Identify the given values. Original length (L) = 2.0 m Cross-sectional area (A) = 1.0 × 10⁻⁶ m² Applied force (F) = 100 N Young's Modulus (Y) = 2.0 × 10¹¹ N/m² Step 2: Recall the formula for Young's Modulus. Y = (F/A) / (ΔL/L) We need to find ΔL (elongation). Step 3: Rearrange the formula to solve for ΔL. ΔL = (F × L) / (A × Y) Step 4: Substitute the values and calculate. ΔL = (100 N × 2.0 m) / (1.0 × 10⁻⁶ m² × 2.0 × 10¹¹ N/m²) ΔL = 200 / (2.0 × 10⁵) ΔL = 100 × 10⁻⁵ ΔL = 1.0 × 10⁻³ m Final Answer: The elongation of the wire is 1.0 × 10⁻³ m (or 1 mm).
• Example 2: Calculating Volumetric Strain and Change in Volume A solid sphere of radius 0.1 m is subjected to a uniform pressure of 1.0 × 10⁵ Pa. If the bulk modulus of the material is 2.0 × 10¹⁰ N/m², calculate the volumetric strain and the change in the volume of the sphere. Step 1: Identify the given values. Radius (r) = 0.1 m Pressure (P) = 1.0 × 10⁵ Pa Bulk Modulus (K) = 2.0 × 10¹⁰ N/m² Step 2: Calculate the original volume of the sphere. V = (4/3)πr³ = (4/3) × 3.14159 × (0.1 m)³ = (4/3) × 3.14159 × 0.001 m³ ≈ 4.189 × 10⁻³ m³ Step 3: Use the formula for Bulk Modulus to find volumetric strain (ΔV/V). K = P / (ΔV/V) (ΔV/V) = P / K (ΔV/V) = (1.0 × 10⁵ Pa) / (2.0 × 10¹⁰ N/m²) = 0.5 × 10⁻⁵ = 5.0 × 10⁻⁶ The volumetric strain is 5.0 × 10⁻⁶. Step 4: Calculate the change in volume (ΔV). ΔV = (ΔV/V) × V ΔV = (5.0 × 10⁻⁶) × (4.189 × 10⁻³ m³) ΔV ≈ 2.0945 × 10⁻⁸ m³ Final Answer: The volumetric strain is 5.0 × 10⁻⁶, and the change in volume is approximately 2.09 × 10⁻⁸ m³.

Exam Traps and Key Tips for Success

To ace your exams in Mechanical Properties of Solids, be mindful of these common pitfalls and essential tips:

1. Unit Consistency is Crucial: Always convert all given quantities to SI units (metres, kilograms, seconds, Newtons, Pascals) before calculation. A mix of units will invariably lead to incorrect answers.
2. Differentiate Stress vs. Pressure: While both are force per unit area and share the same unit, stress is an internal restoring force, whereas pressure is an external applied force. Understand this conceptual difference.
3. Elastic Limit: Remember that Hooke's Law (Stress ∝ Strain) is valid only within the elastic limit. Beyond this limit, the material may deform permanently.
4. Correct Modulus Application: Use Young's Modulus for changes in length, Bulk Modulus for changes in volume, and Shear Modulus for changes in shape. Applying the wrong modulus is a common mistake.
5. Poisson's Ratio Sign: Don't forget the negative sign in the formula for Poisson's ratio, indicating that lateral contraction accompanies longitudinal extension, and vice-versa.
6. Stress-Strain Curve: Be able to draw and interpret the stress-strain curve for ductile and brittle materials, identifying the proportional limit, elastic limit, yield point, ultimate tensile strength, and fracture point. These points provide valuable information about a material's behavior.

Practice Questions with Solutions

• Q: A copper wire of length 2.2 m and cross-sectional area 0.5 × 10⁻⁶ m² is stretched by a load of 40 N. Calculate the stress, strain, and elongation produced in the wire. (Given Young's modulus for copper = 1.1 × 10¹¹ N/m²). A: Step 1: Calculate Stress. Stress (σ) = F/A = 40 N / (0.5 × 10⁻⁶ m²) = 8.0 × 10⁷ N/m² Step 2: Calculate Strain. From Hooke's Law, Y = Stress / Strain => Strain = Stress / Y Strain = (8.0 × 10⁷ N/m²) / (1.1 × 10¹¹ N/m²) ≈ 7.27 × 10⁻⁴ (dimensionless) Step 3: Calculate Elongation (ΔL). Strain = ΔL / L => ΔL = Strain × L ΔL = (7.27 × 10⁻⁴) × 2.2 m ≈ 1.60 × 10⁻³ m Final answer: Stress = 8.0 × 10⁷ N/m², Strain ≈ 7.27 × 10⁻⁴, Elongation ≈ 1.60 × 10⁻³ m.
• Q: A cube of aluminum of side 0.1 m is subjected to a tangential force of 1000 N applied on its upper face, while its lower face is fixed. If the shear modulus of aluminum is 2.5 × 10¹⁰ Pa, calculate the lateral displacement of the upper face. A: Step 1: Identify given values. Side (L) = 0.1 m, Area (A) = L² = (0.1 m)² = 0.01 m² Tangential Force (F_tangential) = 1000 N Shear Modulus (G) = 2.5 × 10¹⁰ Pa Step 2: Calculate Tangential Stress. Tangential Stress = F_tangential / A = 1000 N / 0.01 m² = 1.0 × 10⁵ Pa Step 3: Calculate Shear Strain (θ). G = Tangential Stress / Shear Strain => Shear Strain (θ) = Tangential Stress / G θ = (1.0 × 10⁵ Pa) / (2.5 × 10¹⁰ Pa) = 4.0 × 10⁻⁶ radians Step 4: Calculate Lateral Displacement (Δx). Shear Strain (θ) = Δx / L => Δx = θ × L Δx = (4.0 × 10⁻⁶) × 0.1 m = 4.0 × 10⁻⁷ m Final answer: The lateral displacement of the upper face is 4.0 × 10⁻⁷ m.
• Q: What is the significance of the elastic limit in a stress-strain curve? Explain with respect to material behavior. A: Step 1: Define elastic limit. The elastic limit is the maximum stress a material can withstand without undergoing permanent deformation. Beyond this point, the material will not return to its original shape once the deforming force is removed. Step 2: Explain its significance. For stresses below the elastic limit, the material exhibits elastic behavior, meaning it recovers its original shape completely. Once the stress exceeds the elastic limit, the material enters the plastic region, and even after removing the load, some deformation remains, leading to permanent changes in its structure. This limit is crucial for designing structures and components to ensure they operate within their safe, elastic range. Final answer: The elastic limit defines the boundary where a material transitions from purely elastic behavior to plastic deformation. Operating within this limit ensures a material regains its original shape, which is vital for structural integrity and component longevity.
• Q: A body of mass 200 kg is attached to one end of a wire of length 5 m and diameter 0.02 m. If Young's modulus of the wire is 2.0 × 10¹¹ N/m², find the increase in its length. (Take g = 10 m/s²). A: Step 1: Calculate the force (weight) applied. Force (F) = mass × gravity = 200 kg × 10 m/s² = 2000 N Step 2: Calculate the cross-sectional area (A). Radius (r) = diameter / 2 = 0.02 m / 2 = 0.01 m Area (A) = πr² = 3.14159 × (0.01 m)² = 3.14159 × 10⁻⁴ m² Step 3: Apply Young's Modulus formula to find ΔL. Y = (F/A) / (ΔL/L) => ΔL = (F × L) / (A × Y) ΔL = (2000 N × 5 m) / (3.14159 × 10⁻⁴ m² × 2.0 × 10¹¹ N/m²) ΔL = 10000 / (6.28318 × 10⁷) ΔL ≈ 1.59 × 10⁻⁴ m Final answer: The increase in the length of the wire is approximately 1.59 × 10⁻⁴ m.

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