Reading Time: 2 minutesStress (
Strain (
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 (
)
)The internal resistance per unit area when a force is applied:
![\[ \sigma = \frac{F}{A} \]](https://skinattuition.co.uk/core/ql-cache/quicklatex.com-8a85d53d58b1fb497ea32d3075a3115f_l3.png "Rendered by QuickLaTeX.com")
Where:
- : Stress (Pa or N/m²)
: Applied force (N)
: Cross-sectional area (m²)
: Applied force (N)
: Cross-sectional area (m²)Strain (
)
)The fractional deformation of a material:
![\[ \epsilon = \frac{\Delta L}{L_0} \]](https://skinattuition.co.uk/core/ql-cache/quicklatex.com-08685f1e5116a5453794ea8c17f14ed5_l3.png "Rendered by QuickLaTeX.com")
Where:
- : Unitless strain
: Change in length (m)
: Original length (m)
: Change in length (m)
: Original length (m)Hooke’s Law
The linear relationship between stress and strain in the elastic region:
![\[ \sigma = E\epsilon \]](https://skinattuition.co.uk/core/ql-cache/quicklatex.com-6e9acfe0240dd3b85debb3defacdfe40_l3.png "Rendered by QuickLaTeX.com")
Where 
: Young’s modulus (material stiffness in Pa)
Practical Applications
Structural Engineering
- Steel beams:
- Concrete:
Biomechanics
- Tendons:
- Bone:
Material Science
- Rubber:
- Diamond:
Worked Example
Given:
- Wire length
- Cross-section
- Extension
- Force
- Calculate Stress:
![\[ \sigma = \frac{200}{0.0001} = 2 \, \text{MPa} \]](data:image/svg+xml;base64,PHN2ZyB4bWxucz0iaHR0cDovL3d3dy53My5vcmcvMjAwMC9zdmciIHdpZHRoPSIxNjAiIGhlaWdodD0iMzYiIHZpZXdCb3g9IjAgMCAxNjAgMzYiPjxyZWN0IHdpZHRoPSIxMDAlIiBoZWlnaHQ9IjEwMCUiIHN0eWxlPSJmaWxsOiNjZmQ0ZGI7ZmlsbC1vcGFjaXR5OiAwLjE7Ii8+PC9zdmc+ "Rendered by QuickLaTeX.com")
- Calculate Strain:
![\[ \epsilon = \frac{0.01}{2} = 0.005 \, \text{(0.5\%)} \]](data:image/svg+xml;base64,PHN2ZyB4bWxucz0iaHR0cDovL3d3dy53My5vcmcvMjAwMC9zdmciIHdpZHRoPSIxODQiIGhlaWdodD0iMzYiIHZpZXdCb3g9IjAgMCAxODQgMzYiPjxyZWN0IHdpZHRoPSIxMDAlIiBoZWlnaHQ9IjEwMCUiIHN0eWxlPSJmaWxsOiNjZmQ0ZGI7ZmlsbC1vcGFjaXR5OiAwLjE7Ii8+PC9zdmc+ "Rendered by QuickLaTeX.com")
- Determine Young’s Modulus:
![\[ E = \frac{2 \times 10^6}{0.005} = 400 \, \text{MPa} \]](data:image/svg+xml;base64,PHN2ZyB4bWxucz0iaHR0cDovL3d3dy53My5vcmcvMjAwMC9zdmciIHdpZHRoPSIxODgiIGhlaWdodD0iMzkiIHZpZXdCb3g9IjAgMCAxODggMzkiPjxyZWN0IHdpZHRoPSIxMDAlIiBoZWlnaHQ9IjEwMCUiIHN0eWxlPSJmaWxsOiNjZmQ0ZGI7ZmlsbC1vcGFjaXR5OiAwLjE7Ii8+PC9zdmc+ "Rendered by QuickLaTeX.com")
![\[ \sigma = \frac{200}{0.0001} = 2 \, \text{MPa} \]](https://skinattuition.co.uk/core/ql-cache/quicklatex.com-1b1b1c7aac1dda5bdeeff706a3dc935b_l3.png "Rendered by QuickLaTeX.com")
![\[ \epsilon = \frac{0.01}{2} = 0.005 \, \text{(0.5\%)} \]](https://skinattuition.co.uk/core/ql-cache/quicklatex.com-20a35517f786b52fca31acf05b4b967d_l3.png "Rendered by QuickLaTeX.com")
![\[ E = \frac{2 \times 10^6}{0.005} = 400 \, \text{MPa} \]](https://skinattuition.co.uk/core/ql-cache/quicklatex.com-3609e3ff151f870759c245c751ebde0d_l3.png "Rendered by QuickLaTeX.com")
Common Errors
- Using diameter instead of area in stress calculations
- Confusing engineering strain with true strain
- Applying Hooke’s Law beyond the proportional limit
Practice Problems
- A 1.5m polymer rod (
) stretches 3mm under 150N load. Calculate . - Compare the stiffness of steel () and aluminum ().
- Explain why elastomers have low Young’s modulus values.
) stretches 3mm under 150N load. Calculate 
) and aluminum (
).