Elasticity: Exploring Stress, Strain, and Hooke’s Law in A-Level Science

What Is Elasticity?

Elasticity describes a material’s ability to deform under stress and return to its original shape when the stress is removed, within its elastic limit.

Fundamental Concepts

Stress (\(\sigma\))

The internal resistance per unit area when a force is applied:
\[
\sigma = \frac{F}{A}
\]
Where:

• \(\sigma\): Stress (Pa or N/m²)
• \(F\): Applied force (N)
• \(A\): Cross-sectional area (m²)

Strain (\(\epsilon\))

The fractional deformation of a material:
\[
\epsilon = \frac{\Delta L}{L_0}
\]
Where:

• \(\epsilon\): Unitless strain
• \(\Delta L\): Change in length (m)
• \(L_0\): Original length (m)

Hooke’s Law

The linear relationship between stress and strain in the elastic region:
\[
\sigma = E\epsilon
\]
Where \(E\): Young’s modulus (material stiffness in Pa)

Practical Applications

Structural Engineering

• Steel beams: \(E \approx 200 \, \text{GPa}\)
• Concrete: \(E \approx 30 \, \text{GPa}\)

Biomechanics

• Tendons: \(E \approx 1.5 \, \text{GPa}\)
• Bone: \(E \approx 18 \, \text{GPa}\)

Material Science

• Rubber: \(E \approx 0.01-0.1 \, \text{GPa}\)
• Diamond: \(E \approx 1220 \, \text{GPa}\)

Worked Example

Given:

• Wire length \(L_0 = 2 \, \text{m}\)
• Cross-section \(A = 0.0001 \, \text{m}^2\)
• Extension \(\Delta L = 0.01 \, \text{m}\)
• Force \(F = 200 \, \text{N}\)

1. Calculate Stress:
   \[
   \sigma = \frac{200}{0.0001} = 2 \, \text{MPa}
   \]
2. Calculate Strain:
   \[
   \epsilon = \frac{0.01}{2} = 0.005 \, \text{(0.5\%)}
   \]
3. Determine Young’s Modulus:
   \[
   E = \frac{2 \times 10^6}{0.005} = 400 \, \text{MPa}
   \]

Common Errors

1. Using diameter instead of area in stress calculations
2. Confusing engineering strain with true strain
3. Applying Hooke’s Law beyond the proportional limit

Practice Problems

1. A 1.5m polymer rod (\(A = 5 \times 10^{-5} \, \text{m}^2\)) stretches 3mm under 150N load. Calculate \(E\).
2. Compare the stiffness of steel (\(E = 200 \, \text{GPa}\)) and aluminum (\(E = 69 \, \text{GPa}\)).
3. Explain why elastomers have low Young’s modulus values.