# Elastic and plastic deformation

> CIE A-Level Physics · 9702
> Source: https://www.owlsprep.com/study/cie-9702-u6-elastic-and-plastic-deformation/

This sub-topic explains the two core types of deformation in solid materials under external stress. You will learn to distinguish elastic and plastic behaviour, interpret stress-strain curves, and calculate permanent extension for exam questions.

**Prerequisites:** [Hooke's law and elastic behaviour](https://www.owlsprep.com/study/cie-9702-u6-hookes-law/); [Stress, strain and Young's modulus](https://www.owlsprep.com/study/cie-9702-u6-stress-strain-young-modulus/)

## Learning objectives

- Distinguish between elastic and plastic deformation of solid materials
- Explain the atomic-scale causes of each type of deformation
- Interpret stress-strain graphs to identify deformation regions
- Calculate permanent extension/plastic strain after unloading
- Recognise common exam traps related to this topic

## Elastic Deformation

**Elastic Deformation** — A temporary change in the shape or size of a material when a load is applied. The material returns completely to its original dimensions once all external load is removed.

*Example:* A stretched rubber band that shrinks back to its original length when released

At the atomic level, elastic deformation occurs when inter-atomic bonds stretch temporarily under load. Atoms are displaced from their equilibrium positions, but no bonds break, so atoms return to their original positions once the load is removed. Up to the proportional limit, elastic deformation follows Hooke's law ($F = kx$), where extension is proportional to applied force.

**Worked example:** A spring of original length 15 cm is stretched to 18 cm by a 10 N force, which is below the spring's elastic limit. Describe the length of the spring once the force is completely removed.

1. Since the applied force is below the elastic limit, all deformation is elastic.
2. By definition, elastic deformation is fully reversed when load is removed.
3. The spring will return to its original length of 15 cm.

## Plastic Deformation

**Plastic Deformation** — A permanent change in the shape or size of a material when load is applied above the yield point. The material does not return to its original dimensions once the load is removed.

*Example:* A bent copper wire that stays bent after you release it

Plastic deformation occurs when applied stress exceeds the material's yield point (the stress where plastic flow begins). In crystalline metals, this happens when dislocations allow layers of atoms to slide past each other. Original inter-atomic bonds break, and new bonds form in new positions, so atoms do not return to their original equilibrium after unloading. This leaves a permanent change in shape, called permanent set.

**Worked example:** An aluminium wire of original length 3.0 m is loaded until it yields. After removing all load, its final length is 3.21 m. Calculate the permanent extension of the wire.

1. Deformation beyond the yield point is plastic, so permanent extension remains after unloading.
2. Permanent extension = Final length - Original length
3. $$= 3.21 \, \text{m} - 3.0 \, \text{m} = 0.21 \, \text{m}$$
4. The permanent extension of the wire is 0.21 m (21 cm).

**Check your understanding**

Test your understanding of the difference between the two deformation types

1. Which of the following describes plastic deformation?

   - A: A compressed spring returns to its original shape when released
   - B: A piece of clay that stays squashed after being pressed
   - C: A wooden ruler bent slightly and returning to straight
   - D: A trampoline mat bouncing back after someone jumps on it

   *Why:* Correct: Clay remains permanently deformed after pressing, so this is plastic deformation. All other options return to their original shape, so they are elastic.

## Deformation on Stress-Strain Curves

On a typical stress-strain curve for a ductile material like mild steel: 
1. Deformation is fully elastic from the origin up to the elastic limit
2. Once stress exceeds the yield point, plastic deformation begins
3. When unloading from a point in the plastic region, the unloading line is always parallel to the original linear elastic (Hooke's law) section of the loading curve
4. The permanent strain is the intercept of the unloading line with the strain axis

**Worked example:** A steel sample is loaded to a total strain of 0.015. The stress at maximum load is 2400 MPa, and Young's modulus of steel is 200 GPa. Calculate the permanent strain after unloading.

1. When unloading from plastic deformation, the elastic strain is recovered along a line parallel to the original elastic curve.
2. Calculate elastic strain recovered using $E = \frac{\sigma}{\varepsilon}$:
3. $$\varepsilon_{\text{elastic}} = \frac{\sigma}{E} = \frac{2400 \times 10^6 \, \text{Pa}}{200 \times 10^9 \, \text{Pa}} = 0.012$$
4. Permanent strain = Total strain at maximum load - Recovered elastic strain:
5. $$\varepsilon_{\text{permanent}} = 0.015 - 0.012 = 0.003$$
6. Final permanent strain is 0.003.

*Calculator:* allowed

## Common pitfalls

- **Wrong:** Confusing elastic limit with proportional limit
  - Why it fails: The proportional limit is where Hooke's law stops applying, but elastic deformation continues up to the higher elastic limit.
  - Correct: Remember the order: proportional limit < elastic limit < yield point. Elastic deformation ends at the elastic limit, not the proportional limit.
- **Wrong:** Assuming any deformation beyond the proportional limit is plastic
  - Why it fails: Many materials have a region between the proportional limit and elastic limit where deformation is still fully reversible (elastic), just not proportional to stress.
  - Correct: Deformation is only plastic if it does not return to original shape after unloading, regardless of Hooke's law.
- **Wrong:** Drawing the unloading line back to the origin from plastic deformation
  - Why it fails: Unloading from the plastic region does not follow the original loading curve. The unloading line is always parallel to the initial elastic linear section.
  - Correct: Always draw unloading lines from plastic deformation parallel to the original elastic line to find the correct permanent strain.
- **Wrong:** Taking maximum extension as permanent extension
  - Why it fails: Even when loaded into the plastic region, some elastic extension is still recovered when the load is removed.
  - Correct: Permanent extension = total extension at maximum load minus the elastic extension recovered during unloading.

## Cheatsheet

| Property | Elastic Deformation | Plastic Deformation |
| --- | --- | --- |
| Returns to original shape after unloading? | Yes | No |
| Occurs for stress below: | Elastic limit | N/A (always above yield point) |
| Produces permanent set? | No | Yes |
| Unloading line from max load | Follows loading curve | Parallel to original elastic line |

## What's next

Elastic and plastic deformation is a core concept for the deformation of solids unit in CIE A-Level Physics, and regularly appears in multiple choice and structured questions. Understanding this behaviour underpins further topics like calculating energy stored in deformed solids, describing fracture behaviour of brittle and ductile materials, and applying material properties to real-world engineering problems. Mastery of this topic will also help you interpret stress-strain graphs correctly, which is a common exam requirement.

- [Strain energy](https://www.owlsprep.com/study/cie-9702-u6-strain-energy/)
- [Waves](https://www.owlsprep.com/study/cie-9702-u7-overview/)
- [Progressive waves](https://www.owlsprep.com/study/cie-9702-u7-progressive-waves/)

---

From [OwlsPrep](https://www.owlsprep.com) — free study guides for A-Level, IB, AP and IGCSE, written against the official syllabus. Canonical page: https://www.owlsprep.com/study/cie-9702-u6-elastic-and-plastic-deformation/