recall and use EP = 21 Fx = 21 kx2 for a material deformed within its limit of proportionality

6.2 Elastic & Plastic Behaviour

6.2.1 Learning Objectives

• Recall and use the elastic‑potential‑energy formula $$E_{\text P}= \frac12 F x = \frac12 k x^{2}$$ for a material deformed within its limit of proportionality.
• Explain the difference between the limit of proportionality, the elastic limit, and the yield point/strength.
• Describe a simple experiment (hanging‑mass method) to determine Young’s modulus of a wire.
• Interpret stress–strain and force–extension graphs and identify the regions described above.

6.2.2 Key Definitions

Stress (σ)	Force per unit area $$\sigma = \frac{F}{A}\qquad\text{units: Pa (N m⁻²)}$$
Strain (ε)	Relative extension (dimensionless) $$\varepsilon = \frac{x}{L_{0}}$$
Young’s Modulus (E)	Material constant linking stress and strain in the linear region $$\sigma = E\,\varepsilon \qquad\text{(E in Pa)}$$
Spring constant (k)	Force–extension constant for a linear elastic element $$F = kx$$
Relation between k and E	For a uniform wire of original length L₀ and cross‑sectional area A $$k = \frac{EA}{L_{0}}$$ (derived from σ = Eε and F = σA)
Limit of proportionality	Highest stress at which σ and ε remain strictly proportional – the straight‑line portion of the stress–strain curve.
Elastic limit	Greatest stress at which the material returns completely to its original dimensions when the load is removed. It may lie slightly beyond the limit of proportionality.
Yield point / Yield strength	Stress at which permanent (plastic) deformation begins. In many metals the yield point coincides with the elastic limit, but the syllabus treats them as separate concepts.
Plastic region	Stress beyond the elastic limit where deformation is permanent.

6.2.3 From Force–Extension to Stress–Strain

• Because σ = F/A and ε = x/L₀, a force–extension graph can be transformed into a stress–strain graph by dividing the vertical axis by the cross‑sectional area A and the horizontal axis by the original length L₀.
• The linear (Hookean) region of both graphs is identical in shape; only the scales differ.

6.2.4 Limits on the Linear Region

The diagram below (sketch) is the typical stress–strain curve for a metal:

• Blue straight line: limit of proportionality (Hooke’s law exact).
• Solid red line: elastic limit – material still returns to original shape.
• Dotted red line: yield point / yield strength – onset of permanent deformation.
• Curved region to the right: plastic deformation.

*(In a classroom worksheet the sketch can be drawn with the four markers clearly labelled.)*

6.2.5 Elastic Potential Energy

Within the linear elastic region the force varies linearly with extension, so the work done (and the energy stored) is the area under the force–extension line:

$$E_{\text P}= \int_{0}^{x}F\,dx = \int_{0}^{x}kx\,dx = \frac12 kx^{2}$$

Because F = kx, the same expression can be written as

$$E_{\text P}= \frac12 F x$$

This energy is fully recoverable **provided the material is not loaded beyond its elastic limit**.

For stresses **beyond the elastic limit** the area under the curve represents energy that is dissipated as heat or used to create permanent defects; it is **not** stored as recoverable elastic potential energy.

6.2.6 Experimental Determination of Young’s Modulus (Hanging‑Mass Method)

1. Clamp a uniform wire of known length L₀ and cross‑sectional area A vertically.
2. Attach a set of masses m to the lower end and record the extension x for each mass (measure with a ruler or a vernier calliper).
3. Calculate the applied force for each mass: F = mg (use g = 9.81 m s⁻²).
4. Plot F against x. The straight‑line portion gives the slope k.
5. Obtain Young’s modulus from the relation k = EA/L₀ → $$E = \frac{kL_{0}}{A}$$

Key experimental checks:

• All extensions used must lie within the linear region (below the limit of proportionality).
• Repeat measurements to minimise random error and calculate an average slope.

6.2.7 Worked Example – Steel Wire

Given: \(L_{0}=1.00\;\text{m}\), \(A=2.0\times10^{-6}\;\text{m}^{2}\), \(E=2.0\times10^{11}\;\text{Pa}\), extension \(x=2.0\;\text{mm}=2.0\times10^{-3}\;\text{m}\).

Step	Calculation	Result
1. Spring constant	\(k = \dfrac{EA}{L_{0}} = \dfrac{(2.0\times10^{11})(2.0\times10^{-6})}{1.00}\)	\(k = 4.0\times10^{5}\;\text{N m}^{-1}\)
2. Force	\(F = kx = (4.0\times10^{5})(2.0\times10^{-3})\)	\(F = 8.0\times10^{2}\;\text{N}\)
3. Elastic potential energy	\(E_{\text P}= \tfrac12 kx^{2}= \tfrac12(4.0\times10^{5})(2.0\times10^{-3})^{2}\)	\(E_{\text P}= 0.80\;\text{J}\)

6.2.8 Revision Checklist

• Confirm that the extension (or strain) lies within the linear region before using \(k = EA/L_{0}\) or \(E_{\text P}=½kx^{2}\).
• Stress units = Pa (N m⁻²); strain is dimensionless.
• Limit of proportionality = end of exact Hooke’s law; elastic limit may be slightly further; yield point marks the start of permanent deformation.
• All energy stored up to the elastic limit is recoverable; the area beyond represents dissipated energy.
• When converting between force–extension and stress–strain graphs, divide the vertical axis by A and the horizontal axis by L₀.

6.2.9 Common Misconceptions

1. “All deformation is elastic.” – Only deformation up to the elastic limit is fully recoverable.
2. “The area under any stress–strain curve gives stored energy.” – It gives recoverable elastic energy only up to the elastic limit; beyond that the area represents energy loss.
3. “Hooke’s law applies to any material.” – It is valid only within the limit of proportionality for linear‑elastic materials.
4. “Stress has units of N m⁻² and strain has units of mm mm⁻¹.” – Stress has units of Pa; strain is a pure ratio and has no units.
5. “k = EA/L₀ is the same as E.” – \(k\) is a property of a particular specimen (depends on its geometry); \(E\) is an intrinsic material property.

6.2.10 Summary

Within the linear (elastic) region a material obeys Hooke’s law in two equivalent forms:

$$\sigma = E\varepsilon \qquad\text{or}\qquad F = kx,\; k=\frac{EA}{L_{0}}$$

The work done in stretching the material is stored as elastic potential energy

$$E_{\text P}= \frac12 F x = \frac12 k x^{2}$$

If the load exceeds the elastic limit (or the yield point), permanent deformation occurs and the simple energy formula no longer applies; the extra area under the stress–strain curve represents energy dissipated as heat or used to create defects.

Understanding these limits, being able to read the stress–strain diagram, and knowing how to determine \(E\) experimentally are essential skills for Cambridge IGCSE/A‑Level Physics.