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Engineering · Mechanics of Materials

Stress and Strain Calculator

Find axial stress, strain, and the implied elastic modulus from force and dimensions.

N
Applied axial load.
mm²
Area carrying the load.
m
Length before loading.
mm
Elongation under the load.
Try a scenario
Normal stress (σ)
100MPa

σ = F ÷ A. 1 MPa = 1 N/mm² = 1×10⁶ Pa.

0.0005
strain ε (0.05%)
200
Young’s modulus E (GPa)
Stress vs strain — the slope is Young’s modulus
Linear stress–strain relationship in the elastic region: stress rises in direct proportion to strain, with slope equal to Young’s modulus E100 MPa00ε = 0.0005

Normal stress is force over area, σ = F ÷ A, and strain is the fractional stretch, ε = ΔL ÷ L₀. A 10 kN load on a 100 mm² rod gives σ = 10000 ÷ 0.0001 = 100 MPa. If that 1 m rod stretches 0.5 mm, ε = 0.0005, so the implied modulus is E = σ ÷ ε = 200 GPa.

What stress and strain measure

When you pull on a bar, stress (σ) is how hard the internal material is working — the force spread over the cross-sectional area it acts on, measured in pascals or, more usefully for engineering, megapascals (MPa). Strain (ε) is the material’s response: how much it stretches relative to its original length. Because it is a length divided by a length, strain is a pure number with no units, often written as a percentage. Divide stress by strain and you recover the material’s stiffness — its Young’s modulus (E).

σ = F ÷ A  ·  ε = ΔL ÷ L₀

σ = stress (Pa), F = force (N), A = area (m²); ε = strain (unitless), ΔL = change in length, L₀ = original length

Worked example

A steel rod of 100 mm² cross-section, 1 m long, carries a 10 kN axial pull and stretches 0.5 mm. Find its stress, strain, and the implied modulus.

  1. 1
    Convert the area to m². 100 mm² × 1×10⁻⁶ = 0.0001 m². Working in SI base units keeps the stress in pascals.
  2. 2
    Compute the normal stress. σ = F ÷ A = 10000 N ÷ 0.0001 m² = 1×10⁸ Pa = 100 MPa.
  3. 3
    Convert the elongation to metres. ΔL = 0.5 mm × 1×10⁻³ = 0.0005 m.
  4. 4
    Compute the strain. ε = ΔL ÷ L₀ = 0.0005 ÷ 1 = 0.0005, or 0.05%. Strain is dimensionless.
  5. 5
    Find the implied Young’s modulus. E = σ ÷ ε = 1×10⁸ ÷ 0.0005 = 2×10¹¹ Pa = 200 GPa — consistent with steel.

Stress units and strain

Stress has pressure units; strain is a bare ratio. These conversions let you move between them cleanly.

QuantityValue / relationNotes
1 MPa1 N/mm² = 1×10⁶ PaMost convenient unit for structural stress
1 GPa1000 MPa = 1×10⁹ PaTypical scale for Young’s modulus
Strain εΔL ÷ L₀ (unitless)A ratio; 0.01 strain = 1%
Steel E≈ 200 GPaReference stiffness for structural steel
Aluminium E≈ 69 GPaAbout a third as stiff as steel

Engineering vs true stress, and the elastic limit

This calculator reports engineering stress — force divided by the original cross-sectional area. It is the standard quantity for design and matches published material properties. True stress instead uses the instantaneous area, which shrinks as a specimen necks near failure, so true stress climbs above engineering stress at large deformations. For the small strains of everyday design work the two are practically identical.

The relation E = σ ÷ ε only holds in the elastic region, where stress is proportional to strain (Hooke’s law, σ ∝ ε) and the material springs back on unloading. That linearity ends at the yield point: beyond it the material deforms permanently and a single modulus no longer describes it. Keep applied stress well below yield — and below the yield stress divided by your factor of safety — for a part to behave predictably.

How do I convert cross-sectional area from mm² to m²?
Multiply by 1×10⁻⁶, because there are 1,000,000 square millimetres in a square metre. So 100 mm² becomes 0.0001 m². Using m² and newtons gives stress directly in pascals.
Why is 1 MPa equal to 1 N/mm²?
A megapascal is 1×10⁶ pascals, and one pascal is 1 N/m². Since 1 m² = 1×10⁶ mm², dividing 1×10⁶ N by 1×10⁶ mm² gives exactly 1 N/mm². That equivalence makes MPa the natural unit when areas are quoted in mm².
Is strain unitless?
Yes. Strain is a change in length divided by the original length, so the units cancel and it is a pure number. It is often written as a percentage (ε = 0.0005 = 0.05%) or in microstrain (×10⁻⁶) for very small values.
What is the difference between engineering stress and true stress?
Engineering stress uses the original cross-sectional area; true stress uses the actual, instantaneous area, which shrinks under tension. They match closely at small strains but diverge near failure, where necking reduces the real area and pushes true stress higher.
When is the implied Young’s modulus valid?
Only in the elastic region, where σ ∝ ε (Hooke’s law). E = σ ÷ ε is meaningful while the material still returns to its original length on unloading. Past the yield point the deformation is permanent and a single modulus no longer applies.
What is a typical stress at which steel yields?
Mild structural steel typically yields around 250 MPa, with higher-strength grades reaching 350 MPa or more. The 100 MPa in the worked example sits safely in the elastic region for such steel.