How to use this calculator
- Enter or choose find.
- Enter or choose axial force F.
- Enter or choose cross-sectional area A.
- Enter or choose original length L₀.
- The answer updates as you type. Use Copy result or Share to keep it.
Method
σ = F ÷ A; ε = ΔL ÷ L₀; E = σ ÷ ε; ΔL = F L₀ ÷ (A E)
- F
- axial force
- A
- original cross-sectional area
- L₀
- original length
- ΔL
- change in length
- σ
- engineering stress
- ε
- engineering strain
- E
- Young’s modulus
Worked example
- σ = 10,000 N ÷ 100 mm² = 100 MPa.
- ε = 0.5 ÷ 1,000 = 0.0005 (500 µε).
- E = 100 MPa ÷ 0.0005 = 200,000 MPa = 200 GPa.
A 1 m bar of 100 mm² that stretches 0.5 mm under 10 kN has E = 200 GPa, typical of steel.
What the result means
A modulus near 200 GPa suggests steel, about 70 GPa aluminium, and around 110 GPa copper or titanium alloys. If stress is above the yield strength, the linear relationship no longer holds and the result is not meaningful.
Assumptions and limits
- Uniaxial tension or compression in the linear-elastic range; engineering (original-area) stress and strain.
- Beyond the proportional limit, and for polymers or rubber, stress and strain are not proportional.
Questions
Does it work in compression?
Yes for short members in the elastic range, using the shortening as ΔL. Long members may buckle first.
Why is strain dimensionless?
It is a length divided by a length. Microstrain (µε) is strain × 10⁶.