SeriesCalc Logo

SeriesCalc

🏗️ Stress & Strain Calculator

Calculate stress, strain, and Young's modulus of a material under axial load.

Inputs

Results

Stress (σ)
0.00e+0
psi
Strain (ε)
0.00e+0
dimensionless
Young's Modulus (E)
0.00e+0
psi

Deformation Visualization

FL₀ΔL

Calculator Description

Stress and strain are the two most fundamental quantities in mechanics of materials. Stress is the internal force per unit area, while strain is the fractional elongation relative to the original length. Their ratio, Young's modulus (E), is an intrinsic property describing the material's stiffness.

What this calculator finds

Enter the axial load F, cross-sectional area A, original length L₀ and change in length ΔL to obtain the tensile stress σ, strain ε and Young's modulus E — quantifying the elastic behavior of the material.

Why it matters

  • Assessing safety factor and yield condition of structural parts
  • Predicting deflection and designing for stiffness
  • Comparing stiffness and elasticity when selecting materials
  • Base input for interpreting stress–strain diagrams

Formula

Stress, Strain & Young's Modulus

Within the linear elastic range (Hooke's law), stress is proportional to strain, and the proportionality constant is Young's modulus E.

σ=FA\sigma = \dfrac{F}{A}
ε=ΔLL0\varepsilon = \dfrac{\Delta L}{L_0}
E=σε=FL0AΔLE = \dfrac{\sigma}{\varepsilon} = \dfrac{F L_0}{A \Delta L}
  • FApplied axial force [N or lbf]
  • ACross-sectional area [m² or in²]
  • L₀Original length [m or in]
  • ΔLChange in length [m or in]
  • σ, ε, EStress [Pa], strain [dimensionless], Young's modulus [Pa]

How the formula works

  • Larger F raises stress σ; larger area A lowers it.
  • Strain ε is proportional to ΔL and inversely proportional to L₀.
  • A larger E means a "stiffer" material that stretches less under the same load (steel > aluminum > rubber).

Worked example

For a 10 mm diameter rod (A ≈ 78.5 mm² = 7.85×10⁻⁵ m²) and original length 1 m, a 10 kN (1×10⁴ N) tensile load causing 0.5 mm elongation gives: σ = 1×10⁴ / 7.85×10⁻⁵ ≈ 127 MPa, ε = 0.5×10⁻³ / 1 = 5×10⁻⁴, and E ≈ 127×10⁶ / 5×10⁻⁴ ≈ 254 GPa — close to steel.

Useful Tips

Practical tips

  • Use nominal (engineering) stress F/A; beyond necking, distinguish true stress.
  • If area varies, the maximum stress occurs at the minimum section A_min — design against that.
  • E generally decreases slightly as temperature rises.

Limitations & cautions

  • Valid only within the elastic limit; beyond yield, plastic deformation not described by E occurs.
  • Buckling, shear and torsion are different deformation modes not covered here.
  • Under multiaxial stress the simple proportionality breaks; apply a yield criterion.