How to calculate a rod’s axial extension
A uniform rod stretches when pulled. This formula combines its load, length, area and stiffness to estimate the small elastic change in length.
What the formula is saying
Stress is N/A, strain is stress/E, and extension is strain × L. Joining these three ideas gives ΔL = NL/(AE). A longer rod stretches more; a larger area or modulus reduces extension.
Read the symbols in plain language
- N
- Axial forceN
- L
- Member lengthm
- A
- Aream²
- E
- Young’s modulusPa
Sort out the units first
Use N, m, m² and Pa together. For this example, 50 kN = 50,000 N and 200 GPa = 200,000,000,000 Pa. Convert the final metres to millimetres by multiplying by 1,000.
Let’s solve one together
Read the given values, follow each operation, then check what the result means.
- N · Axial force
- 50000 N
- L · Member length
- 2 m
- A · Area
- 0.001 m²
- E · Young’s modulus
- 200000000000 Pa
Calculate axial stress
Start with force per cross-sectional area.
(50000) ÷ (0.001) = 50000000 PaConvert stress to strain
Hooke’s law gives strain = stress / stiffness.
(50000000) ÷ (200000000000) = 0.00025Turn the ratio into a length change
Multiply the strain by the original rod length.
(0.00025) × (2) = 0.0005 m
Does this worked answer make sense?
The worked extension is 0.5 mm, much smaller than the 2 m starting length. Doubling A would halve the extension.
Now try your own values
Change a value or its unit. The same method will show your calculation, step by step.
Results update only when you calculate. The lesson example above stays unchanged.
Your turn — check your understanding
Use these new values. Work it out first, then check your answer.
- N · Axial force
- 30000 N
- L · Member length
- 3 m
- A · Area
- 0.0015 m²
- E · Young’s modulus
- 200000000000 Pa
Find: a rod’s axial extension
A hint, not the answer
Stress is N/A, strain is stress/E, and extension is strain × L. Joining these three ideas gives ΔL = NL/(AE). A longer rod stretches more; a larger area or modulus reduces extension.
Use N, m, m² and Pa together. For this example, 50 kN = 50,000 N and 200 GPa = 200,000,000,000 Pa. Convert the final metres to millimetres by multiplying by 1,000.
Show the full practice solution
Compare the steps with your work; revealing a solution does not mark the lesson complete.
Calculate axial stress
Start with force per cross-sectional area.
(30000) ÷ (0.0015) = 20000000 PaConvert stress to strain
Hooke’s law gives strain = stress / stiffness.
(20000000) ÷ (200000000000) = 0.0001Turn the ratio into a length change
Multiply the strain by the original rod length.
(0.0001) × (3) = 0.0003 m
Avoid the common trap
Putting E in GPa while treating it as Pa gives a huge false extension. Using total surface area instead of cross-sectional area also breaks the model.
When this method applies — and when it does not
Constant axial force, uniform cross-section and modulus, small deformation and linear elasticity. For changing sections or loads, solve appropriate segments or use integration.
For study and understanding, not approval of a real structure, site operation or design. Apply the correct standard, National Annex and professional review to actual engineering work.
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Sources & further reading
References open in a new tab and explain the underlying principles. The teaching text and examples here are SimpleFlick’s own; the source organisations have not endorsed this calculator.
Lesson updated: · Worked examples checked against the implemented formula; not an independent engineering certification.