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6. Deformation of solids

Syllabus
9702–2028–2029
Section
6
Level
AS

6.1 Stress and strain

Syllabus
9702–2028–2029
Topic
6.1
Level
AS

Tensile forces stretch a body and compressive forces shorten or squeeze it

A tensile force pulls along a body and tends to extend it; a compressive force pushes along it and tends to shorten or squash it.

Identify the force direction relative to the body axis, then decide whether the deformation is extension or compression. Real materials can experience both locally.

A hanging cable is under tension, while a column supporting a roof is under compression.

“Tension” is not a synonym for any force in a material; it specifically describes pulling stress along the relevant direction.

Load, extension, compression and limit of proportionality describe how a material responds

Load is the applied force; extension is the increase in length and compression the decrease. The limit of proportionality is where extension ceases to be directly proportional to load.

Read these terms from a force–extension graph and distinguish proportional behaviour from later elastic or plastic behaviour.

If a spring extends 2 mm under 4 N and remains proportional, 8 N predicts 4 mm; beyond the limit that scaling may fail.

The limit of proportionality is not automatically the breaking point or the elastic limit; those are separate boundaries.

Hooke’s law states that extension is proportional to force up to the limit of proportionality

For a spring or wire within its proportional region, F=kx: extension x is directly proportional to applied force F.

The relation applies only while the material remains proportional. Use the original length and the extension, not the final length.

If 6 N produces 3 mm extension in the proportional region, 10 N predicts 5 mm for the same spring.

Hooke’s law does not hold indefinitely; crossing the limit of proportionality makes the simple linear relation unreliable.

The spring constant k measures stiffness through k=F/x

The spring constant is k=F/x in the proportional region. Its unit is N m⁻¹; a larger k means more force is needed for the same extension.

Convert extension to metres before calculating. Compare like-for-like loading conditions and do not use points beyond proportional behaviour.

A 12 N force producing 0.040 m extension gives k=300 N m⁻¹.

k is not the extension itself, and a stiffer spring has a larger k, not a larger extension under the same force.

Stress is force per area, strain is extension per original length, and Young modulus links them

Stress σ=F/A and strain ε=x/L are dimensionless ratios of deformation. Young modulus E=σ/ε measures a material’s stiffness in the linear elastic region.

Use cross-sectional area and original length, not the deformed dimensions, and keep stress in pascals.

A wire’s stress doubles if the same force acts on half the area; its strain depends on extension relative to its starting length.

Strain has no unit, while stress and Young modulus are measured in Pa; Young modulus is not the same as spring constant.

Determine Young modulus from a loaded metal wire

Clamp a long straight metal wire securely beside a scale. Measure original test length L between the fixed point and a marker. Measure diameter d with a micrometer at several positions and in perpendicular orientations; average d and calculate A = πd²/4.

Add known masses in small steps, allowing the wire to settle. For each load calculate F = mg and measure extension x from the marker. Repeat readings where practical and remain below the limit of proportionality.

Plot F against x for the straight-line region. Its gradient is F/x. Since E = FL/(Ax), calculate E = (L/A) × gradient. Equivalently, the gradient of stress against strain is E.

Use a long thin wire to make x measurable, avoid parallax, and average diameter because A depends on d². Wear eye protection and shield/catch falling masses; do not load near breaking.

6.2 Elastic and plastic behaviour

Syllabus
9702–2028–2029
Topic
6.2
Level
AS

Elastic deformation is reversible, while plastic deformation leaves a permanent change

Elastic deformation disappears when the load is removed; plastic deformation remains. The elastic limit is the greatest load or stress before permanent deformation begins.

Separate the elastic limit from the limit of proportionality: a material may stop being linear before it becomes permanently deformed.

A metal wire unloaded within its elastic region returns to its original length; loaded beyond its elastic limit it remains longer.

Elastic does not mean perfectly linear, and plastic does not mean the object has already broken.

Force–extension area represents work done

During a small extension Δx at approximately constant force F, work done is FΔx. Adding these narrow strips over the loading path gives the total area under the force–extension graph.

Force is in newtons and extension in metres, so graph area has unit N m = J. For straight segments, calculate and add rectangle, triangle or trapezium areas; for a curve, estimate the area under it.

If force rises linearly from 10 N to 30 N while extension increases by 0.20 m, the trapezium area is ½(10 + 30) × 0.20 = 4.0 J of work.

Final force × total extension is the whole rectangle and is generally not the work for a changing force. Follow the actual loading graph and its axis order.

Elastic potential energy is the proportional loading area

When a material is loaded within its limit of proportionality, deformation is reversible and the work done is stored as elastic potential energy. On a force–extension graph, this is the area under the loading line.

The proportional line runs from the origin to final point (x, F), so the area is a triangle: elastic potential energy = ½ × extension × final force.

A spring reaches 40 N at extension 0.20 m on a straight line from the origin. Stored energy = ½ × 0.20 × 40 = 4.0 J.

This triangular-area result requires loading within the limit of proportionality. For a non-linear graph, determine the actual area rather than assuming a triangle.

For a linear spring, elastic potential energy is ½Fx or ½kx²

Within the proportional region, E_P=½Fx=½kx² because the average force during loading is F/2.

Use final force and extension only for a straight-line force–extension graph from zero. For a non-linear graph, find the area under the curve.

A spring with k=200 N m⁻¹ stretched 0.10 m stores 1.0 J.

Do not use Fx for a spring whose force changes during loading; that would overestimate the work by a factor of two in the linear case.