understand and use the terms load, extension, compression and limit of proportionality

Stress, Strain and the Limit of Proportionality

Learning Objectives

• Define load (force), extension, compression and the limit of proportionality (LoP).
• Distinguish between tensile and compressive loading.
• Apply Hooke’s law to calculate stress, strain, extension and the spring constant for a rod.
• Read and interpret a stress–strain graph: locate the linear (elastic) region, LoP, elastic limit, yield point, ultimate tensile strength and fracture.
• Explain how Young’s modulus relates to the slope of the linear region and to material stiffness.
• Identify key experimental considerations (area measurement, safety limits, avoiding permanent deformation).

Key Definitions

Fundamental Formulas

• Stress: \$\displaystyle \sigma = \frac{F}{A}\$ (where \$A\$ is the original cross‑sectional area).
• Strain: \$\displaystyle \varepsilon = \frac{\Delta L}{L_{0}}\$.
• Hooke’s law (elastic region): \$\displaystyle \sigma = E\,\varepsilon\$.
• Spring‑constant form for a straight rod (derived from Hooke’s law):
  

  \$\displaystyle F = k\,\Delta L\qquad\text{with}\qquad k = \frac{EA}{L_{0}}\$.

Derivation of \$k = EA/L_{0}\$

1. Start with \$\sigma = E\varepsilon\$.
2. Replace \$\sigma\$ by \$F/A\$ and \$\varepsilon\$ by \$\Delta L/L_{0}\$:
   

   \$\displaystyle \frac{F}{A}=E\frac{\Delta L}{L_{0}}\$.

3. Re‑arrange to obtain \$F = \left(\frac{EA}{L_{0}}\right)\Delta L\$; the term in brackets is the spring constant \$k\$ for the rod.

Typical Values of Young’s Modulus

Stress–Strain Graph (Cambridge A‑Level)

Graph‑Reading Exercise

1. Estimate the stress at the LoP (\$\sigma_{\text{LoP}}\$) from the vertical axis.
2. Using the data in the example below, calculate the stress produced by the given load. Compare it with \$\sigma_{\text{LoP}}\$ and state whether the load lies within the linear region.
3. Read the strain at the yield point. Comment on the difference between the LoP and the yield point for a ductile metal such as steel.

Worked Example – Elastic Extension of a Steel Rod

Given:

• Original length \$L_{0}=1.20\ \text{m}\$
• Diameter \$d=1.80\ \text{cm}\$ → \$A=\dfrac{\pi d^{2}}{4}=2.54\times10^{-4}\ \text{m}^{2}\$
• Young’s modulus \$E=2.0\times10^{11}\ \text{Pa}\$
• Tensile load \$F=5.0\times10^{4}\ \text{N}\$ (assumed to be within the elastic region)

1. Stress: \$\displaystyle \sigma=\frac{F}{A}= \frac{5.0\times10^{4}}{2.54\times10^{-4}}=1.97\times10^{8}\ \text{Pa}\$.
2. Strain (Hooke’s law): \$\displaystyle \varepsilon=\frac{\sigma}{E}= \frac{1.97\times10^{8}}{2.0\times10^{11}}=9.85\times10^{-4}\$ (≈ 0.098 %).
3. Extension: \$\displaystyle \Delta L = \varepsilon L_{0}= (9.85\times10^{-4})(1.20)=1.18\times10^{-3}\ \text{m}=1.18\ \text{mm}\$.
4. From the typical steel curve, \$\sigma{\text{LoP}}\approx2.0\times10^{8}\ \text{Pa}\$. Since \$\sigma=1.97\times10^{8}\ \text{Pa}<\sigma{\text{LoP}}\$, the load is safely within the linear region.

Distinguishing Tensile and Compressive Loading

• Tensile loading: pulls the material, giving a positive extension (\$\Delta L>0\$) and a positive stress (\$\sigma>0\$).
• Compressive loading: pushes the material, giving a negative extension (\$\Delta L<0\$) and a negative stress (\$\sigma<0\$). In practice the magnitude of compressive stress is quoted as a positive number, but the loading type must be noted.
• Both obey \$\sigma = E\varepsilon\$ while the material remains in the elastic region.

Common Misconceptions – Clarified

• Load vs. Stress: Load is the total force \$F\$ (N). Stress is the force per unit area (\$\sigma = F/A\$) and has units of pressure (Pa).
• Extension vs. Strain: Extension \$\Delta L\$ is an absolute change in length (m). Strain \$\varepsilon\$ is the relative change (\$\Delta L/L_{0}\$) and is dimensionless.
• LoP vs. Yield Point: LoP marks the end of the proportional (linear) relationship. The yield point marks the onset of permanent deformation; they may be close for many metals but are distinct concepts.
• Units: Stress → Pa (or N mm⁻²). Strain → dimensionless (often expressed as a %). They are not interchangeable.

Typical Linear‑Elastic Strain Ranges

• Steel: up to ≈ 0.20 % strain (\$2\times10^{-3}\$) before noticeable deviation.
• Aluminium: up to ≈ 0.15 % strain.
• Copper: up to ≈ 0.25 % strain.
• Polymers (e.g., PMMA): up to 1 %–2 % strain.

Experimental & Safety Considerations

• Measure the cross‑sectional area accurately (calipers for diameter, then \$A=\pi d^{2}/4\$ for circular rods; \$A=wh\$ for rectangular bars).
• Apply loads gradually; record extension after each increment to avoid sudden fracture.
• Never exceed the ultimate tensile strength shown on the stress–strain curve – failure is catastrophic.
• When using a universal testing machine, set the maximum load just below the estimated \$\sigma_{\text{LoP}}\$ for a first trial.
• Always wear safety goggles and keep a safe distance from the testing apparatus.

Symbol Summary