Chapter 8: Mechanical Properties of Solids
Unit VII – Properties of Bulk Matter
8.1 Elastic Behaviour of Solids
Solids resist deformation when forces are applied. They return to their original shape after the force is removed — this is called elasticity.
- Elastic body: Returns to original shape (steel spring)
- Plastic body: Does not return (putty, clay)
8.2 Stress
Stress is the restoring force per unit area:
\[ \sigma = \frac{F}{A} \]
Types of stress:
| Type | Description | Formula |
|---|---|---|
| Tensile/Compressive | Along length | \(\sigma = F/A\) |
| Shear | Lateral (parallel to surface) | \(\tau = F/A\) |
| Volumetric | All-around pressure | \(p = F/A\) |
SI unit: Pascal (Pa) = N/m²
8.3 Strain
Strain is the fractional change in dimension:
\[ \epsilon = \frac{\Delta L}{L} \]
Types:
| Type | Formula | Notes |
|---|---|---|
| Longitudinal strain | \(\Delta L / L\) | Change in length |
| Shear strain | \(\tan\phi \approx \phi\) | Angle of shear |
| Volumetric strain | \(\Delta V / V\) | Change in volume |
Strain has no unit (dimensionless).
8.4 Hooke’s Law
Within the elastic limit, stress is directly proportional to strain.
\[ \text{Stress} \propto \text{Strain} \quad \Rightarrow \quad \text{Stress} = E \times \text{Strain} \]
where \(E\) is the modulus of elasticity.
8.5 Elastic Moduli
Young’s Modulus (Y)
For longitudinal stress/strain:
\[ Y = \frac{\text{Tensile stress}}{\text{Longitudinal strain}} = \frac{F/A}{\Delta L/L} = \frac{FL}{A\Delta L} \]
Bulk Modulus (B)
For volumetric stress/strain:
\[ B = -\frac{\Delta p}{\Delta V/V} \]
(negative sign: volume decreases as pressure increases)
Compressibility \(= 1/B\)
Shear Modulus (G) (Modulus of Rigidity)
\[ G = \frac{\text{Shear stress}}{\text{Shear strain}} = \frac{F/A}{\phi} \]
8.6 Stress-Strain Curve
Points on the curve:
- O → A: Linear (Hooke’s law applies); slope = Young’s modulus
- A: Proportional limit (end of linear behaviour)
- B: Elastic limit (body still returns to original shape)
- C: Yield point (onset of plastic deformation)
- D: Ultimate strength (maximum stress before necking)
- E: Fracture point
8.7 Poisson’s Ratio
When a rod is stretched longitudinally, it contracts laterally:
\[ \sigma = -\frac{\text{lateral strain}}{\text{longitudinal strain}} = -\frac{\Delta D/D}{\Delta L/L} \]
Range: \(-1 \leq \sigma \leq 0.5\) (most materials: 0.2–0.4)
8.8 Elastic Potential Energy
Energy stored per unit volume in a stretched wire:
\[ u = \frac{1}{2} \times \text{stress} \times \text{strain} = \frac{Y(\text{strain})^2}{2} \]
Total elastic energy:
\[ U = \frac{1}{2}F\Delta L \]
Key Formulas Summary
| Formula | Quantity |
|---|---|
| \(Y = \dfrac{FL}{A\Delta L}\) | Young’s modulus |
| \(B = -\dfrac{p}{\Delta V/V}\) | Bulk modulus |
| \(\text{compressibility} = 1/B\) | Compressibility |
| \(\sigma = -\dfrac{\Delta D/D}{\Delta L/L}\) | Poisson’s ratio |
| \(U = \dfrac{1}{2}F\Delta L\) | Elastic PE |
Practice Questions
Section A – MCQ (1 mark each)
Q1. The Young’s modulus has the same units as:
(a) Strain (b) Stress (c) Force (d) Energy
Answer
(b) Stress — \(Y = \dfrac{\text{stress}}{\text{strain}}\), strain is dimensionless, so \([Y] = [\text{stress}] = \text{Pa}\)
Q2. A wire of length 2 m and cross-section area \(10^{-6}\) m² is stretched by 2 mm by a force of 100 N. Young’s modulus is:
(a) \(10^{11}\) Pa (b) \(10^{10}\) Pa (c) \(10^9\) Pa (d) \(2 \times 10^{11}\) Pa
Answer
(b) \(10^{10}\) Pa
\(Y = \dfrac{FL}{A\Delta L} = \dfrac{100 \times 2}{10^{-6} \times 2 \times 10^{-3}} = \dfrac{200}{2 \times 10^{-9}} = 10^{11} \text{ Pa}\)
Wait — recalculating: \(Y = \dfrac{100 \times 2}{10^{-6} \times 0.002} = \dfrac{200}{2 \times 10^{-9}} = 10^{11} \text{ Pa}\). Answer: (a)
Section B – Short Answer (2–3 marks)
Q3. Explain the stress-strain curve for a metallic wire and identify the elastic and plastic regions.
Answer
- O to A (elastic region): Stress proportional to strain (Hooke’s law). Slope = Young’s modulus.
- A = proportional limit; B = elastic limit (body returns to original shape if stress removed)
- B to E (plastic region): Permanent deformation occurs.
- C = yield point (large strain with little stress increase)
- D = ultimate tensile strength (maximum stress)
- E = fracture point (wire breaks)
Section D – Competency-Based Questions
Q4. (Case Study) A steel wire of length 4 m has a cross-sectional diameter of 2 mm. When a load of 400 N is applied: (\(Y_{steel} = 2 \times 10^{11} \text{ Pa}\))
(i) Calculate the elongation.
(ii) Calculate the stress and strain.
(iii) What is the elastic PE stored in the wire?
Answer
\(A = \pi r^2 = \pi \times (10^{-3})^2 = \pi \times 10^{-6} \text{ m}^2 \approx 3.14 \times 10^{-6} \text{ m}^2\)
(i) \(\Delta L = \dfrac{FL}{AY} = \dfrac{400 \times 4}{3.14 \times 10^{-6} \times 2 \times 10^{11}} = \dfrac{1600}{6.28 \times 10^5} \approx \mathbf{2.55 \times 10^{-3} \text{ m}}\)
(ii) Stress \(= \dfrac{F}{A} = \dfrac{400}{3.14 \times 10^{-6}} \approx 1.27 \times 10^8 \text{ Pa}\)
Strain \(= \dfrac{\Delta L}{L} = \dfrac{2.55 \times 10^{-3}}{4} \approx 6.37 \times 10^{-4}\)
(iii) \(U = \frac{1}{2}F\Delta L = \frac{1}{2} \times 400 \times 2.55 \times 10^{-3} \approx \mathbf{0.51 \text{ J}}\)