Deformation of Solids - A-Level Physics

📚 Based on Cambridge A-Level Physics Syllabus - Topics 6.1 & 6.2

6.1 Stress and Strain

Syllabus Points: Tensile/Compressive forces, Hooke's Law, Stress, Strain, Young's Modulus

1. Types of Deformation

Deformation occurs when forces act on a solid material, changing its shape or size. Forces can be:
  • Tensile Forces: Stretching forces that increase length
  • Compressive Forces: Squeezing forces that decrease length

2. Key Terms

Load (F): Force applied to a material (N)
Extension (x): Increase in length under tensile load (m)
Compression: Decrease in length under compressive load (m)
Limit of Proportionality: Point beyond which extension is no longer proportional to load

3. & 4. Hooke's Law and Spring Constant

Hooke's Law: The extension of a spring is directly proportional to the load applied, provided the limit of proportionality is not exceeded.
F = kx
Where:
• F = Force applied (N)
• k = Spring constant (N/m)
• x = Extension (m)
Spring Constant (k): A measure of stiffness. Higher k = stiffer spring.

🔬 Example 1: Spring Extension

A spring has a spring constant of 200 N/m. Calculate the extension when a force of 50 N is applied.

Solution:

F = kx → 50 = 200 × x

x = 50 ÷ 200 = 0.25 m (25 cm)

5. Stress, Strain and Young's Modulus

Stress (σ): Force per unit cross-sectional area
σ = F / A
Strain (ε): Extension per unit original length
ε = x / L
Young's Modulus (E): Ratio of stress to strain - measures material stiffness
E = σ / ε = (F × L) / (A × x)
Units:
• Stress (σ): Pascals (Pa) or N/m²
• Strain (ε): No units (dimensionless)
• Young's Modulus (E): Pascals (Pa)

📐 Example 2: Young's Modulus Calculation

A steel wire of length 2.0 m and diameter 0.5 mm extends by 1.0 mm under a load of 50 N. Calculate Young's Modulus.

Solution:

Cross-sectional area A = πr² = π × (0.25×10⁻³)² = 1.96×10⁻⁷ m²

Stress σ = F/A = 50 ÷ (1.96×10⁻⁷) = 2.55×10⁸ Pa

Strain ε = x/L = 0.001 ÷ 2.0 = 0.0005

Young's Modulus E = σ/ε = 2.55×10⁸ ÷ 0.0005 = 5.10×10¹¹ Pa

6. Experiment: Determining Young's Modulus

🧪 Young's Modulus Experiment for Metal Wire

Apparatus:
  • Long thin wire of the metal
  • Micrometer screw gauge
  • Meter rule
  • Masses and hanger
  • Vernier scale or travelling microscope
  • Safety frame
Procedure:
  1. Measure original length L of wire
  2. Measure diameter at several points using micrometer
  3. Calculate average cross-sectional area A = πr²
  4. Add masses gradually and measure extension x
  5. Plot graph of force F against extension x
  6. Calculate gradient = F/x
  7. Young's Modulus E = (gradient × L) / A
Safety: Use safety frame to catch weights if wire breaks

6.2 Elastic and Plastic Behaviour

Syllabus Points: Elastic/Plastic deformation, Force-extension graphs, Elastic potential energy

1. Types of Deformation Behaviour

Elastic Deformation: Material returns to original shape when load is removed
Plastic Deformation: Material remains permanently deformed when load is removed
Elastic Limit: Point beyond which material will not return to original shape

2. & 3. Force-Extension Graphs and Work Done

📈 Typical Force-Extension Graph

Limit of Proportionality
Elastic Limit
Fracture Point
Force (N)
Extension (m)
Area under force-extension graph = Work done in deforming the material
For materials obeying Hooke's Law (within limit of proportionality), the area is a triangle

4. Elastic Potential Energy

Elastic Potential Energy: Energy stored in a deformed material within its elastic limit
Ep = ½Fx = ½kx²
Derivation:
Work done = Area under F-x graph
For Hookean material: Area = ½ × base × height = ½ × x × F
Since F = kx, substitute: Ep = ½ × x × kx = ½kx²

💡 Example 3: Elastic Potential Energy

A spring with k = 100 N/m is compressed by 0.1 m. Calculate the energy stored.

Solution:

Ep = ½kx² = ½ × 100 × (0.1)² = 0.5 J

📊 Example 4: Work Done from Graph

A force-extension graph shows a straight line up to 5 N force and 0.02 m extension. Calculate work done.

Solution:

Area under graph = ½ × 0.02 × 5 = 0.05 J

Material Properties Comparison

Steel
E ≈ 200 GPa
Copper
E ≈ 110 GPa
Aluminium
E ≈ 70 GPa
Rubber
E ≈ 0.01 GPa

Key Relationships Summary

Hooke's Law: F = kx
Stress: σ = F/A
Strain: ε = x/L
Young's Modulus: E = σ/ε = FL/(Ax)
Elastic Energy: Ep = ½Fx = ½kx²

Exam Tips

🎯 Graph Skills: Be able to interpret force-extension graphs and identify key points
📐 Units Matter: Always convert mm to m, cm to m before calculations
🧮 Area Calculations: Remember area under F-x graph = work done
🔍 Experimental Details: Know the Young's modulus experiment procedure
✅ These notes comprehensively cover ALL syllabus requirements for Cambridge A-Level Physics Topics 6.1 and 6.2