Example: 1mm² = 1e-6 m²

Deformation of Solids

Materials undergo deformation when subject to external forces. The Young Modulus (E) is a numerical constant that describes a material's inherent stiffness. It is independent of the object's dimensions and depends only on the material itself (e.g., steel vs. rubber).

E = Tensile Stress (σ) / Tensile Strain (ε)

1. Tensile Stress (σ)

Stress is the internal force per unit area acting on a material. It represents how "hard" the particles are being pulled apart.

σ = Force / Area = F / A

2. Tensile Strain (ε)

Strain is the fractional change in length. Because it is a ratio of two lengths, it has no units.

ε = Extension / Original Length = e / L

3. Elastic Potential Energy

When a material is stretched within its elastic limit, it stores energy. On a Force-Extension graph, this energy is represented by the area under the graph.

Energy = ½ × Force × Extension
Exam Tip: Don't confuse the Young Modulus with the Spring Constant (k). The Spring Constant (k = F/e) describes a specific object (like a particular spring), while the Young Modulus describes a material (like Copper).

Deep Dive: Worked Examples

✅ Example 1: Basic Young Modulus Calculation

A 2.0m long wire with area 1.0 mm² extends by 2.0 mm under a 100 N load. Calculate E.

Step 1: Convert units A = 1.0 × 10⁻⁶ m², e = 2.0 × 10⁻³ m
Step 2: Calculate Stress and Strain σ = 100 / 1e-6 = 1.0 × 10⁸ Pa.
ε = 2e-3 / 2.0 = 1.0 × 10⁻³.
Step 3: Solve E = 1e8 / 1e-3 = 1.0 × 10¹¹ Pa (100 GPa).

✅ Example 2: Force for a specific extension

Steel has E = 200 GPa. Find the force needed to stretch a 1.5m steel rod (diameter 2mm) by 0.5mm.

Step 1: Calculate Area A = πr² = π(1e-3)² ≈ 3.14 × 10⁻⁶ m²
Step 2: Rearrange E = (FL) / (Ae) F = (E × A × e) / L
Step 3: Solve F = (200e9 × 3.14e-6 × 0.5e-3) / 1.5 ≈ 209.4 N

✅ Example 3: Strain Energy Stored

Calculate the energy stored in the wire from Example 1.

Step 1: Use E = ½Fe F = 100 N, e = 2.0 × 10⁻³ m
Step 2: Multiply E = 0.5 × 100 × 0.002
Step 3: Solve E = 0.10 J

✅ Example 4: Comparing Two Wires

Wire A and Wire B are made of the same material. Wire B is twice as long and has twice the diameter. Compare their extensions under the same load.

Step 1: Identify relationships e = FL / AE. Diameter × 2 → Area × 4.
Step 2: Compare parameters e_B = F(2L) / (4A)E = 0.5 (FL/AE)
Step 3: Conclusion Wire B will extend half as much as Wire A.

✅ Example 5: Yield Point Stress

A material fails at a stress of 4.5e8 Pa. If the wire has a diameter of 0.8mm, find the breaking force.

Step 1: Area A = π(0.4e-3)² ≈ 5.03e-7 m²
Step 2: Force = Stress × Area F = 4.5e8 × 5.03e-7
Step 3: Solve F ≈ 226 N