Elastic and plastic behaviour

1. Overview

Deformation occurs when a force is applied to an object, causing a change in its shape or size. In A-Level Physics, we analyze this through the relationship between the applied force (load) and the resulting extension. The fundamental principle is that work must be done to stretch or compress a material; this work is transferred into the material's internal energy. If the deformation is reversible, the energy is stored as elastic potential energy ($E_P$). If the deformation is permanent, the work done is dissipated as thermal energy. Understanding the transition from elastic to plastic behaviour is critical for engineering and material science, as it determines the structural integrity and energy-storage capacity of materials.

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Key Definitions

• Elastic Deformation: A change in the shape or size of an object where the object returns to its original dimensions (length, area, or volume) when the deforming force (load) is removed.
• Plastic Deformation: A change in the shape or size of an object that is permanent; the object does not return to its original dimensions when the deforming force is removed.
• Elastic Limit: The maximum force or stress that can be applied to a material such that it will still return to its original dimensions when the force is removed. Beyond this point, plastic deformation occurs.
• Limit of Proportionality: The point beyond which extension is no longer directly proportional to the applied force. This is the point where Hooke’s Law ($F \propto x$) ceases to be true.
• Elastic Potential Energy ($E_P$): The energy stored in a body due to its change in shape (deformation), which is recovered when the deforming force is removed. It is equal to the work done to deform the material elastically.
• Extension ($x$ or $\Delta L$): The difference between the extended length ($L$) and the original length ($L_0$) of a material ($x = L - L_0$).
• Spring Constant ($k$): The force per unit extension for a material obeying Hooke's Law, representing the "stiffness" of the object. Measured in $\text{N},\text{m}^{-1}$.

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Content

3.1 Elastic and Plastic Behaviour

The macroscopic behaviour of a material is determined by the microscopic interactions between its atoms.

• The Atomic Basis of Elasticity: In a solid, atoms are held in an equilibrium position by interatomic forces. When a tensile force is applied, the atoms are pulled slightly further apart. As long as the displacement is small, the atoms will return to their equilibrium positions once the force is removed. This is elastic behaviour.
• The Atomic Basis of Plasticity: If the applied force exceeds the elastic limit, the atoms are displaced so far that they "slide" over one another into new equilibrium positions. When the force is removed, the atoms do not return to their original positions, but stay in their new arrangement. This results in a permanent extension.
• Loading and Unloading Curves:
  • Elastic Region: The loading and unloading curves follow the same path on a Force-Extension graph.
  • Plastic Region: If a material is stretched into the plastic region and then the load is removed, the unloading curve will be a straight line parallel to the initial linear (Hooke's Law) region. The point where this unloading line intersects the x-axis (extension axis) represents the permanent extension (or permanent set).

3.2 Work Done and the Force-Extension Graph

When you stretch a material, you are moving a force through a distance. This requires work.

• Variable Force: Unlike moving a block at a constant speed, the force required to stretch a spring increases as the extension increases. Therefore, we cannot simply use $W = F \times d$ with a single value of $F$.
• Area Under the Graph: For any Force-Extension ($F-x$) graph, the area under the graph represents the total work done on the material.
  • Within the Elastic Limit: All work done is stored as elastic potential energy ($E_P$). This energy is fully recoverable.
  • Beyond the Elastic Limit: The area under the loading curve is the total work done. The area under the unloading curve is the energy recovered. The area between the loading and unloading curves (the hysteresis loop) represents the work done to permanently deform the material, which is dissipated as thermal energy (heat).

3.3 Derivation of Elastic Potential Energy

For a material that obeys Hooke’s Law ($F = kx$), the relationship between force and extension is linear.

1. Definition of Work: Work done ($W$) is the area under the $F-x$ graph.
2. Geometric Approach: For a material within the limit of proportionality, the $F-x$ graph is a straight line passing through the origin, forming a triangle.
3. Area Calculation: $$\text{Area of triangle} = \frac{1}{2} \times \text{base} \times \text{height}$$ $$\text{Work Done} (W) = \frac{1}{2} \times x \times F$$
4. Energy Identity: Since this work is stored as potential energy: $E_P = \frac{1}{2}Fx$
5. Substitution using Hooke's Law: Substitute $F = kx$ into the equation: $$E_P = \frac{1}{2}(kx)x$$ $E_P = \frac{1}{2}kx^2$

Note: These equations only apply if the material is within its limit of proportionality.

3.4 Worked Examples

Worked Example 1 — Energy Stored in a Stretched Wire

A steel wire of original length $2.50 , \text{m}$ is stretched by a force of $80.0 , \text{N}$, resulting in a new length of $2.52 , \text{m}$. Assuming the wire remains within its limit of proportionality, calculate the elastic potential energy stored.

• Step 1: Calculate the extension ($x$) $x = L - L_0 = 2.52 , \text{m} - 2.50 , \text{m} = 0.02 , \text{m}$
• Step 2: Identify the correct formula Since we have Force ($F$) and extension ($x$), use $E_P = \frac{1}{2}Fx$.
• Step 3: Substitution $E_P = 0.5 \times 80.0 , \text{N} \times 0.02 , \text{m}$
• Step 4: Final Answer $E_P = 0.80 , \text{J}$

Worked Example 2 — Comparing Energy in Springs

Spring A has a spring constant of $400 , \text{N},\text{m}^{-1}$. Spring B has a spring constant of $800 , \text{N},\text{m}^{-1}$. Both springs are compressed by $5.0 , \text{cm}$. Calculate the ratio of energy stored in Spring B to Spring A.

• Step 1: Convert units to SI $x = 5.0 , \text{cm} = 0.050 , \text{m}$
• Step 2: Use $E_P = \frac{1}{2}kx^2$ for both $E_{P(A)} = 0.5 \times 400 \times (0.050)^2 = 200 \times 0.0025 = 0.50 , \text{J}$ $E_{P(B)} = 0.5 \times 800 \times (0.050)^2 = 400 \times 0.0025 = 1.00 , \text{J}$
• Step 3: Calculate the ratio $\text{Ratio} = \frac{E_{P(B)}}{E_{P(A)}} = \frac{1.00}{0.50} = 2$
• Step 4: Conclusion Spring B stores twice the energy of Spring A for the same extension because its spring constant is doubled.

Worked Example 3 — Non-linear Energy Calculation

A polymer strip is stretched. The area under its force-extension loading curve is found to be $15.0 , \text{J}$. When the load is removed, the area under the unloading curve is $6.5 , \text{J}$. Calculate the energy dissipated as heat during this cycle.

• Step 1: Understand the energy balance Total Work Done (Loading) = Elastic Potential Energy (Recovered) + Thermal Energy (Dissipated).
• Step 2: Set up the calculation $\text{Thermal Energy} = \text{Area under loading curve} - \text{Area under unloading curve}$
• Step 3: Substitution $\text{Thermal Energy} = 15.0 , \text{J} - 6.5 , \text{J}$
• Step 4: Final Answer $\text{Energy dissipated} = 8.5 , \text{J}$

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Key Equations

Equation	Symbols	SI Units	Data Sheet?
$F = kx$	$F$: Force, $k$: Spring constant, $x$: Extension	$F$ (N), $k$ ($\text{N},\text{m}^{-1}$), $x$ (m)	No (Must recall)
$E_P = \frac{1}{2}Fx$	$E_P$: Elastic Potential Energy, $F$: Force, $x$: Extension	$E_P$ (J), $F$ (N), $x$ (m)	Yes
$E_P = \frac{1}{2}kx^2$	$E_P$: Elastic Potential Energy, $k$: Spring constant, $x$: Extension	$E_P$ (J), $k$ ($\text{N},\text{m}^{-1}$), $x$ (m)	Yes
$W = \text{Area under } F-x \text{ graph}$	$W$: Work done	$W$ (J)	No (Conceptual)

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Common Mistakes to Avoid

• ❌ Wrong: Using the total length of the wire/spring in the formula $E_P = \frac{1}{2}kx^2$. ✅ Right: Always calculate the extension ($x = \text{new length} - \text{original length}$) before substituting.
• ❌ Wrong: Forgetting to square the extension in $\frac{1}{2}kx^2$. This is the most common calculation error in Topic 6. ✅ Right: Write the formula out clearly and perform the squaring step separately if needed.
• ❌ Wrong: Assuming $E_P = \frac{1}{2}kx^2$ applies to the plastic region. ✅ Right: This formula is derived from the area of a triangle (linear proportionality). If the graph is curved, you must use the area under the curve (e.g., by counting squares).
• ❌ Wrong: Confusing the Elastic Limit with the Limit of Proportionality. ✅ Right: The Limit of Proportionality is where the graph stops being a straight line. The Elastic Limit is where the material stops being able to return to its original shape. In many materials, these points are very close, but they are conceptually different.
• ❌ Wrong: Mixing units, such as using $k$ in $\text{N},\text{m}^{-1}$ with $x$ in $\text{mm}$. ✅ Right: Convert all lengths to meters (m) immediately. $1 , \text{mm} = 10^{-3} , \text{m}$.

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Exam Tips

1. Graph Interpretation: If a question provides a graph, check the axes carefully. If it is an Extension-Force graph (axes swapped), the area under the graph is still work done, but the gradient will be $1/k$ rather than $k$.
2. Counting Squares: When calculating the area under a non-linear graph:
   • Find the "value" of one small square: $\text{Value} = (\text{Force per unit}) \times (\text{Extension per unit})$.
   • Count the total number of full squares.
   • Estimate the number of partial squares (usually by combining two halves into one).
   • $\text{Total Energy} = \text{Total Squares} \times \text{Value of one square}$.
3. The "Parallel" Rule: When a material undergoes plastic deformation, the unloading line on an $F-x$ graph is always parallel to the initial linear part of the loading curve. Use a ruler to draw this in exams to find the permanent extension.
4. Significant Figures: Cambridge 9702 papers typically require answers to 2 or 3 significant figures. If the input data is $12 , \text{N}$ and $2.0 , \text{mm}$, provide your answer to 2 s.f.
5. Explain Questions: If asked to explain why a material is "plastic," mention that work is done to move atoms into new positions and this energy is dissipated as heat rather than being stored as $E_P$.
6. Springs in Series/Parallel: While often covered in Topic 6.1, remember that for springs in parallel, $k_{total} = k_1 + k_2$. For springs in series, $1/k_{total} = 1/k_1 + 1/k_2$. You may need to find the effective $k$ before calculating $E_P$.