Edexcel A Level (IAL) Physics-1.28 Stress, Strain & The Young Modulus- Study Notes- New Syllabus
Edexcel A Level (IAL) Physics -1.28 Stress, Strain & The Young Modulus- Study Notes- New syllabus
Edexcel A Level (IAL) Physics -1.28 Stress, Strain & The Young Modulus- Study Notes -Edexcel A level Physics – per latest Syllabus.
Key Concepts:
- 1.28 understand how to use the relationships
• (tensile or compressive) stress = force/cross-sectional area
• (tensile or compressive) strain= change in length/original length
Young modulus = stress/strain
Stress, Strain, and Young Modulus
These quantities describe how materials deform under applied forces and are fundamental in understanding material strength and elasticity.
Tensile and Compressive Stress
Definition: Stress is the force applied per unit cross-sectional area.
$ \text{Stress} = \dfrac{F}{A} $
- \( F \) = applied force (N)
- \( A \) = cross-sectional area (m²)
- Unit of stress: Pascal (Pa) or \( \mathrm{N\,m^{-2}} \)
Types of Stress:
- Tensile stress: when a material is stretched.
- Compressive stress: when a material is squashed.
Tensile and Compressive Strain
Definition: Strain is the fractional change in length of a material.
$ \text{Strain} = \dfrac{\Delta L}{L_0} $
- \( \Delta L \) = change in length (m)
- \( L_0 \) = original length (m)
- Strain has no units (dimensionless).
Types of Strain:
- Tensile strain: increase in length.
- Compressive strain: decrease in length.
Young Modulus (Elastic Modulus)
Definition: Young Modulus describes the stiffness of a material — how much stress is needed to produce a certain strain.
$ E = \dfrac{\text{Stress}}{\text{Strain}} $
- \( E \) = Young Modulus (Pa)
- Large \( E \): very stiff material (e.g., steel).
- Small \( E \): flexible material (e.g., rubber).
Key Notes:
- Young modulus applies in the elastic region where Hooke’s Law holds.
- Beyond the elastic limit, the material does not return to its original length.
- Stress–strain graphs help identify elastic limit, yield point, and breaking stress.
Rearranged Forms
- Stress:
\( \sigma = \dfrac{F}{A} \)
- Strain:
\( \epsilon = \dfrac{\Delta L}{L_0} \)
- Young modulus:
\( E = \dfrac{\sigma}{\epsilon} \)
Example (Easy)
A force of \( 100\, \mathrm{N} \) is applied to a wire of cross-sectional area \( 2.0\times10^{-6}\, \mathrm{m^2} \). Find the stress.
▶️ Answer / Explanation
\( \sigma = \dfrac{F}{A} = \dfrac{100}{2.0\times10^{-6}} = 5.0\times10^{7}\, \mathrm{Pa} \)
Example (Medium)
A wire stretches from \( 1.500\, \mathrm{m} \) to \( 1.503\, \mathrm{m} \) under load. Find the strain.
▶️ Answer / Explanation
Change in length:
\( \Delta L = 1.503 – 1.500 = 0.003\, \mathrm{m} \)
Strain:
\( \epsilon = \dfrac{0.003}{1.500} = 2.0\times10^{-3} \)
Example (Hard)
A steel wire of length \( 2.0\, \mathrm{m} \) and cross-sectional area \( 1.0\times10^{-6}\, \mathrm{m^2} \) is stretched \( 0.50\, \mathrm{mm} \) by a force of \( 500\, \mathrm{N} \). Find the Young Modulus.
▶️ Answer / Explanation
Stress:
\( \sigma = \dfrac{500}{1.0\times10^{-6}} = 5.0\times10^{8}\, \mathrm{Pa} \)
Strain:
\( \Delta L = 0.50\,\mathrm{mm} = 0.00050\, \mathrm{m} \)
\( \epsilon = \dfrac{0.00050}{2.0} = 2.5\times10^{-4} \)
Young Modulus:
\( E = \dfrac{\sigma}{\epsilon} = \dfrac{5.0\times10^{8}}{2.5\times10^{-4}} = 2.0\times10^{12}\, \mathrm{Pa} \)
Young modulus ≈ \( 2.0\times10^{12}\, \mathrm{Pa} \)