recall and use the formula for the spring constant k = F / x

Stress and Strain (AS‑Level 6.1)

Learning Objectives

• Define normal stress (σ) and engineering strain (ε) and state their SI units.
• Explain the limit of proportionality, elastic limit, and the difference between elastic and plastic behaviour.
• State and use Young’s modulus (E = σ/ε) and relate it to the spring constant via k = EA/L₀.
• Calculate the elastic‑potential energy stored in a material or a spring.
• Describe a simple experiment for determining Young’s modulus of a wire (wire‑extension method).
• Apply the formula k = F / x to a range of situations and interpret the result.

Key Quantities & Formulas

Stress–Strain Curve – Key Regions

• Limit of proportionality (LP): σ ∝ ε (Hooke’s law region).
• Elastic limit (EL): maximum stress for which the material returns to its original dimensions when the load is removed.
• Yield point (YP) (ductile materials): onset of permanent (plastic) deformation.
• Fracture point (FP): material breaks.

The gradient of the straight‑line segment LP → EL is Young’s modulus E.

The shaded area under this linear segment represents the elastic‑potential energy per unit volume.

Hooke’s Law, Spring Constant and Their Connection to Young’s Modulus

For a uniform wire of original length L₀, cross‑sectional area A and Young’s modulus E:

\[

F = kx \qquad\text{where}\qquad k = \frac{EA}{L_{0}}

\]

Thus a larger Young’s modulus (stiffer material) or a larger cross‑section produces a larger spring constant.

Elastic‑Potential Energy

• For a coil spring: \(U = \tfrac12 kx^{2}\).
• For any material expressed in stress and strain: \(U = \tfrac12 \sigma\varepsilon V = \tfrac12 E\varepsilon^{2}V\), where V is the volume considered.

Experimental Determination of Young’s Modulus – Wire‑Extension Method

1. Measure the original length L₀ and diameter (or radius) of the wire; calculate cross‑sectional area \(A = \pi d^{2}/4\).
2. Hang a series of known masses m from the wire and record the total length L for each load.
3. For each load calculate:

   • Force: \(F = mg\) (use \(g = 9.81\ \text{m s}^{-2}\)).
   • Extension: \(x = L - L_{0}\).
   • Strain: \(\varepsilon = x / L_{0}\).
   • Stress: \(\sigma = F / A\).

4. Plot σ (y‑axis) against ε (x‑axis). The gradient of the initial straight‑line portion is Young’s modulus E.

Worked Example 1 – Determining a Spring Constant

A vertical spring supports a 0.5 kg mass and stretches by 0.040 m. Find the spring constant k.

1. Force due to the mass: \(F = mg = 0.5 \times 9.81 = 4.905\ \text{N}\).
2. Spring constant: \(k = \dfrac{F}{x} = \dfrac{4.905}{0.040} = 122.6\ \text{N m}^{-1}\).
3. To three significant figures: \(k = 1.23 \times 10^{2}\ \text{N m}^{-1}\).

Worked Example 2 – Young’s Modulus of a Copper Wire

Given: L₀ = 1.20 m, d = 0.80 mm, mass = 2.0 kg, extension = 0.35 mm.

1. Cross‑sectional area: \(A = \dfrac{\pi d^{2}}{4}

   = \dfrac{\pi (0.80\times10^{-3})^{2}}{4}

   = 5.03\times10^{-7}\ \text{m}^{2}\).

2. Force: \(F = mg = 2.0 \times 9.81 = 19.62\ \text{N}\).
3. Stress: \(\sigma = \dfrac{F}{A}

   = \dfrac{19.62}{5.03\times10^{-7}}

   = 3.90\times10^{7}\ \text{Pa}\).

4. Strain: \(\varepsilon = \dfrac{\Delta L}{L_{0}}

   = \dfrac{0.35\times10^{-3}}{1.20}

   = 2.92\times10^{-4}\).

5. Young’s modulus: \(E = \dfrac{\sigma}{\varepsilon}

   = \dfrac{3.90\times10^{7}}{2.92\times10^{-4}}

   = 1.34\times10^{11}\ \text{Pa}\) (≈ 134 GPa, a realistic value for copper).

Common Mistakes & How to Avoid Them

• Using the total length of a spring or wire instead of the change in length (x or ΔL). Always work with the extension/compression only.
• Confusing mass with force. Convert mass to force with \(F = mg\) before using k = F / x.
• Ignoring units. Stress is in pascals (N m⁻²), not newtons; strain is dimensionless.
• Reading the gradient of a curved (plastic) part of the σ‑ε graph as E. Only the initial linear region (LP → EL) is valid.
• Incorrect significant‑figure handling. Your final answer should reflect the precision of the measured quantities.

Practice Questions

1. A spring with \(k = 250\ \text{N m}^{-1}\) is stretched by \(0.06\ \text{m}\). What force is required?
2. A force of \(15\ \text{N}\) compresses a spring by \(0.030\ \text{m}\). Determine the spring constant and state whether this spring is stiffer or softer than the spring in question 1.
3. A mass of \(2.0\ \text{kg}\) hangs from a spring causing an extension of \(0.080\ \text{m}\). Find:

   1. the spring constant, and
   2. the elastic‑potential energy stored in the spring.

4. Using the wire‑extension method, a steel wire (diameter = 0.50 mm, length = 1.00 m) shows a stress of \(2.0\times10^{8}\ \text{Pa}\) when the strain is \(1.0\times10^{-3}\). Calculate Young’s modulus and comment on whether the material is still behaving elastically at this load.
5. Sketch a stress‑strain diagram for a typical ductile metal. Label the limit of proportionality, elastic limit, yield point and fracture point, and shade the area that represents elastic‑potential energy.

Summary

The central idea of AS‑Level stress and strain is the linear relationship between normal stress and engineering strain, expressed by Young’s modulus (E = σ/ε). The limit of proportionality marks the region where Hooke’s law (σ ∝ ε) is valid; the elastic limit is the highest stress for which the material returns to its original shape. The spring constant \(k = F/x\) is a practical measure of stiffness for a coil spring and is linked to material properties by \(k = EA/L_{0}\). Mastery of these concepts enables you to:

• analyse elastic behaviour of solids,
• calculate the energy stored during deformation,
• determine material constants experimentally, and
• compare the stiffness of different springs or wires using the formulae above.