Normal stress
MPa
Positive in tension, negative in compression.
Mechanics · Stress and deformation
Axial stress and bar elongation
Evaluate stress, strain and length change of a prismatic bar under a centred axial force.
Created v1.0
By Thibaut Grzelak, Mechanical Analysis Engineer
03
MetricResults
Axial strain
µε
Dimensionless strain expressed in microstrain.
Length change
mm
Positive elongation or negative shortening.
Axial stiffness
kN/mm
Ratio of axial force to displacement.
View calculation detailsFormulas, SI conversion, numerical substitution and resultsOpen ↓
Calculation note
Full calculation details
Centred axial force and constant section. Homogeneous isotropic linear-elastic material. Small strain. No stress concentration, buckling, plasticity or thermal effect.
01Formulas and symbols
Formulas used
σ
N/ANormal stressε
σ/EAxial strainΔL
NL/(EA)Length changek
EA/LAxial stiffness02Assumptions and limits
Scope of validity
- Centred axial force and constant section.
- Homogeneous isotropic linear-elastic material.
- Small strain.
- No stress concentration, buckling, plasticity or thermal effect.
03Validation example
Reference numerical case
- N = 120 kN, A = 1,500 mm², E = 210 GPa and L = 2 m.
- σ = 80 MPa.
- ε = 380.952 µε.
- ΔL = 0.7619 mm and k = 157.5 kN/mm.
04Frequently asked questions
Questions about the calculation
Which sign convention is used?
A positive force represents tension. A negative force represents compression and gives negative stress and length change.
When is ΔL = NL/(EA) valid?
For a straight prismatic linear-elastic bar under a centred axial force with constant E and A.
Does the calculation cover compression buckling?
No. A slender compressed member must also be checked with the Euler buckling calculator or an applicable design code.
05References
Technical references
- Gere & Goodno, Mechanics of Materials, axial tension and compression.
- Hibbeler, Mechanics of Materials, axial deformation.