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Worked example
Open this modelAxial response
Inputs
- Axial loadF
- 10 kN
- Cross-sectional areaA
- 1000 mm²
- Original lengthL
- 1000 mm
- Elastic modulusE
- 200 GPa
Results
- Average normal stressσ
- 10 MPa
- Elastic strainε
- 50 µε
- Ideal length changeΔL
- 0.05 mm
Relation
σ = F / A · ε = σ / E · ΔL = FL / AE
Assumptions
Uniform cross-section · Axial loading · Linear-elastic response
Where it stops
Not a stress concentration, not a joint, not a code check.
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