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Structural Analysis Fundamentals / Stress and strain

Hooke's law, stiffness and strength

The elastic relationship, and three ideas that are constantly confused.

Lesson review: expert verification remains open

This lesson has received a full-context editorial review and a check of its technical explanations. Selected arithmetic has been checked separately. This is not independent engineering sign-off.

Checks needed for project use

  • An independent structural reviewer must check remaining derivations, diagrams, sign conventions and the interactive models across their full input ranges.
  • Independent review of diagrams, idealisations and full interactive input ranges remains required.

These are educational models and exercises. Check their assumptions before using the results. The full technical content and every diagram have not yet been independently verified.

Concrete and steel calculations retain their stated first-generation Eurocode teaching basis. Use the standards and amendments specified for your project; do not mix generations without an appropriate design basis.

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Why this matters

Material stiffness determines the elastic stress–strain relationship. Member stiffness also depends on geometry, while strength describes a resistance limit. Distinguishing them helps identify which change will address a design problem.

By the end of this lesson you should be able to

  • Apply Hooke's law
  • Calculate axial extension
  • Separate material stiffness, member stiffness and strength
Hooke's law and axial extension
σ=Eεδ=PLAE\sigma = E\varepsilon \qquad \delta = \frac{PL}{AE}

What it calculates: The elastic relationship between stress and strain, and the resulting change in length of an axially loaded member.

E
Young's modulus of the material (N/mm²)
σ
Direct stress (N/mm²)
ε
Direct strain
δ
Extension (positive) or shortening (mm)
P
Axial force (N)
L
Original length (mm)
A
Cross-sectional area (mm²)

This assumes

  • The material is still behaving elastically — below the limit of proportionality
  • The load, area and material are uniform along the member

In plain terms: E is the slope of the straight part of the stress–strain graph. A stiffer material has a steeper slope, so it needs more stress to reach the same strain. Steel is about 205 000 N/mm², aluminium about 70 000, timber roughly 11 000 along the grain.

Worked example

Stress, strain and extension in a steel tie

Given

  • A steel tie carries a tensile force of 180 kN
  • Cross-sectional area 1500 mm²
  • Length 3.2 m
  • E = 205 000 N/mm²

Find

The direct stress, the strain, and how much the tie stretches.

Assumptions

  • The steel stays elastic
  • The section is uniform and away from connections

    Practice

    A steel bar of area 800 mm² carries a tensile force of 96 kN. What is the direct stress?

    Practice

    A steel bar 4.0 m long with a cross-sectional area of 900 mm² carries 135 kN in tension. With E = 205 000 N/mm², how much does it stretch?

    Practice

    What is the axial stiffness AE/L of a member with A = 1200 mm², E = 205 000 N/mm² and L = 3000 mm? Give your answer in kN/mm.

    Practice

    Two uniform bars have the same cross-sectional area, steel modulus and axial load. Bar B is twice as long as bar A. What is the ratio of B’s elastic extension to A’s?

    Summary

    • σ = Eε while the material is elastic
    • δ = PL/AE: extension depends on load, length, area and material
    • Material stiffness E, member stiffness AE/L and strength are three different things
    • Stress concentrations and yielding put a limit on simple theory