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Mechanics
Axial Bar Stress and Extension
Enter values in the stated units. Example inputs are provided to help you explore the method.
Results
Your results will appear here after calculation. Changing an input clears the previous result.
Method & assumptions
- For a uniform bar under centered axial load, stress σ = F/A, strain ε = σ/E, and length change δ = εL.
- Example: 10,000 N, 0.001 m², 2 m and E = 200 GPa gives 10 MPa stress, 0.00005 strain and 0.0001 m extension. The modulus is illustrative; use material- and temperature-appropriate data.
- Negative force represents compression and produces negative strain and shortening in this model.
- Valid only for small linear-elastic deformation. Does not check yield, buckling, fatigue, connections, stress concentrations or eccentric loading. Compression results do not establish column stability.
- Source: https://openstax.org/books/university-physics-volume-1/pages/12-3-stress-strain-and-elastic-modulus