Free · Physics models
Axial Stress, Strain and Elastic Modulus
Relate axial force, cross-section and length change to engineering stress, strain and elastic modulus.
- Formula & worked example
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Calculator inputs
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How to use this calculator
- Enter the known values in the units shown. Results update as you type.
- Where results are editable, change one to solve backwards. Lock a value to hold it fixed.
- Use the worked example to check the method. Reset restores the starting fields.
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Inputs and results stay in this browser tab. Bookify does not upload or store the values you enter.
Formula and method
stress = 0.10000000; strain = 0.010000000; length change = 0.010000000. The worked example lists its inputs and unit conventions.
σ = F/A; ε = (Lf−L0)/L0; E = σ/ε.
Worked example
Enter these known values and leave the other values blank.
- Area (A)
- 0.01 m²
- Force (F)
- 1000 N
- Initial length (L₁)
- 1 m
- Final length (L₂)
- 1.01 m
- Stress (σ)
- 0.1 MPa
- Change in length (ΔL)
- 0.01 m
- Strain (ε)
- 0.01
- Young's modulus (E)
- 0.01 GPa
Assumptions and limitations
- Uniform uniaxial loading and small engineering strains are assumed. Stress and strain may both be negative for compression. Young modulus must be positive and the modulus inverse needs nonzero strain.
- Clear the unknown to solve an inverse and lock quantities that must remain fixed. This does not model buckling, plastic deformation or changes in cross-section.
- Uniform uniaxial, small-strain, linear elasticity. Engineering stress and strain share signs for positive Young modulus. Nonzero strain is required for the finite modulus inverse.
Common questions
Is engineering stress true stress?
No. Engineering stress divides by the original cross-section. True stress uses the current section and requires further information.
References
Young modulus must be positive and strain must be nonzero for the finite modulus inverse.