Elastic Modulus Calculator
Calculate Young’s modulus from tensile test force, cross-section, gauge length and extension, or from stress and strain, in MPa, GPa and psi.
Interactive calculator · Method ELASTIC-MODULUS-1.0 · Last reviewed: 2026-11-04
Calculate your scenario
Stress σ = force ÷ cross-section area (N ÷ mm² = MPa), with area = π d² ÷ 4 for a round bar.
Your inputs
Default worked example. Edit the inputs to calculate your scenario.
Default worked example
σ = 10,000 N ÷ 78.5398 mm² = 127.324 MPa; ε = 0.03 ÷ 50 = 0.0006; E = σ ÷ ε = 212.207 GPa.
How to use the calculator
Choose the method. For test data, enter force, the bar diameter or area, the gauge length and the extension at that force. Otherwise enter stress in MPa and strain in percent.
What the result means
A 10 mm bar under 10 kN carries about 127.324 MPa; stretching 0.03 mm over 50 mm (strain 0.0006) implies E ≈ 212.207 GPa, typical of steel.
Formula and units
Stress σ = force ÷ cross-section area (N ÷ mm² = MPa), with area = π d² ÷ 4 for a round bar. Strain ε = extension ÷ original gauge length. Young’s modulus E = σ ÷ ε; GPa = MPa ÷ 1,000 and psi = MPa ÷ 0.006894757.
Worked examples
10 kN on a 10 mm bar, 0.03 mm over 50 mm: Applied tensile force: 10000 N; Diameter or area: 10 mm or mm²; Original gauge length: 50 mm; Measured extension: 0.03 mm; Stress (stress and strain method): 200 MPa; Strain (stress and strain method): 0.1 %; Input method: force; Cross-section entry: diameter.
Example result: Young’s modulus: 212.207 GPa; Young’s modulus in psi: 30.778 × 10⁶ psi; Stress: 127.324 MPa; Strain: 0.0006; Strain in percent: 0.06%.
70 MPa at 0.1% strain: Applied tensile force: 10000 N; Diameter or area: 10 mm or mm²; Original gauge length: 50 mm; Measured extension: 0.03 mm; Stress (stress and strain method): 70 MPa; Strain (stress and strain method): 0.1 %; Input method: stress; Cross-section entry: diameter.
Example result: Young’s modulus: 70 GPa; Young’s modulus in psi: 10.153 × 10⁶ psi; Stress: 70 MPa; Strain: 0.001; Strain in percent: 0.1%.
Limitations and common mistakes
Assumes linear-elastic, small-strain behavior and uniaxial tension or compression on the original cross-section. The answer is only as good as the test data entered: use force and extension from the linear part of the curve, ideally as a slope between two points. This is not a design allowable or material specification.
Using total crosshead movement instead of extensometer readings over the gauge length; machine compliance makes the extension look larger and the modulus too low.
Sources and assumptions
Method ELASTIC-MODULUS-1.0; last formula review 2026-11-04. Sources and their supported claims. Calculation methodology.