Library / Mechanical / MEC-02
Mechanical
Axial stress and extension
How much a bar stretches under a pull, and the stress in it.
- Ref
- MEC-02
- Rev
- v1
- Field
- Mechanical
- Result
- δ [mm]
δ is the extension, F the axial force, L the original length, A the cross-section area and E Young’s modulus. Stress σ = F/A and strain ε = δ/L.
A bar pulled by a force stretches by an amount proportional to the force and its length, and shrinks with a thicker bar or a stiffer material.
Calculator
- Stress σ = F/A125 MPa
- Strain ε = δ/L625 µε
What drives the result
Each variable is moved across its typical range while the others stay at the values above. The longest bar is the key driver.
- Four inputs act as simple ratios. The force and length push the answer up, the area and stiffness pull it down.
- With these units, MPa equals N/mm², so the stress is easy to compare with the yield stress in the material data.
- Stress, not force alone, decides whether the bar yields. Check the stress against the yield strength with a safety factor.
Chart data as tables
Ranked variables
| Variable | Typical range | Lowest result | Highest result | Swing (mm) | Local step | Change in result |
|---|---|---|---|---|---|---|
| E Young’s modulus | 5 to 400 GPa | 0.625 | 50 | 49.38 | +10% | −9.09 % |
| A Cross-section area | 20 to 5,000 mm² | 0.1 | 25 | 24.9 | +10% | −9.09 % |
| F Axial force | 1 to 500 kN | 0.025 | 12.5 | 12.47 | +10% | +10 % |
| L Length | 0.1 to 20 m | 0.0625 | 12.5 | 12.44 | +10% | +10 % |
Sweep of Axial force
| Axial force (kN) | Extension (mm) |
|---|---|
| 1 | 0.025 |
| 2.175 | 0.05436 |
| 4.729 | 0.1182 |
| 10.28 | 0.2571 |
| 22.36 | 0.559 |
| 48.62 | 1.216 |
| 105.7 | 2.643 |
| 229.9 | 5.748 |
| 500 | 12.5 |
Sweep one variable
Variables and how each one contributes
Extension
What the equation gives you. Everything else feeds it.
Axial force
Force along the axis of the bar. Directly proportional while the material stays elastic.
Length
Original unstretched length. Directly proportional: a longer bar stretches more.
Cross-section area
Original area normal to the force. Inversely proportional: doubling the area halves the extension.
Young’s modulus
Stiffness of the material (steel about 200, aluminium about 70). Inversely proportional: a stiffer material stretches less.
Worked example with the values above
- Stress: σ = F/A = 50,000 N / 400 mm² = 125 MPa.
- Strain: ε = σ/E = 125 / 200,000 = 6.25×10−4.
- Extension: δ = εL = 6.25×10−4 × 2,000 mm = 1.25 mm.
Assumptions and limits
- Uniform bar, constant axial force and constant modulus.
- Linear elastic region (below yield).
- Small strain; original area and length used.
Common mistakes
- Mixing mm² and m²: a factor of 10⁶ error.
- Using the load as stress. Stress is force per area.
- Applying the formula beyond yield.
Statistics and uncertainty
All four inputs enter as a product or ratio, so relative errors combine as the relative error in δ is about √[(δF/F)² + (δL/L)² + (δA/A)² + (δE/E)²]. Young’s modulus of a given steel batch is typically known to ±3% and the area to ±2%.
Where it is used
- Cable, tie-rod and bolt stretch.
- Preload in fasteners.
- Checking elongation of tethers and mast stays.
Maths you need
- Stress and strain: force per area; extension per length
- Units: N/mm² = MPa