Engineer's Notebook §3.1

Axial stress and strain

A bar pulled along its axis carries the force spread evenly over its cross-section. The stress, and the strain it causes, are:

σ = FA,ε = σE
σ
normal stress
Pa
F
axial force
N
A
cross-sectional area
m²
ε
strain
—
E
Young's modulus
Pa
L
original length
m
ΔL
change in length
m
d
bar diameter
m
τ
shear stress
Pa
A round bar fixed to a wall at the left and pulled to the right by a force. The stretched bar is drawn over its original length. F = 20 kN L = 2 m extension drawn ×50
Fig. 3.1-1Extension exaggerated; scale noted under the bar.
A = πd²/4113 mm²
σ = F/A177 MPa
ε = σ/E8.84 × 10⁻⁴
ΔL = FL/AE1.77 mm

Worked example

Uses the values in Fig. 3.1-1. Change them and the working updates.

Given
  • Faxial pull20 kN
  • dbar diameter12 mm
  • Llength2 m
  • Esteel S275200 GPa
Find

The stress in the bar and how much it stretches.

Solution
A = πd²/4 = π × (12 mm)² / 4 = 113 mm² σ = F/A = 20,000 N / 113 mm² = 177 MPa ε = σ/E = 177 MPa / 2 × 10⁵ MPa = 8.84 × 10⁻⁴ ΔL = εL = 8.84 × 10⁻⁴ × 2,000 mm = 1.77 mm
Answer
ΔL = 1.77 mm

The stress is 64.3% of the yield strength of steel S275 (275 MPa).

Related cases

  • Extension

    ΔL = FLAE

  • Pin in single shear

    τ = FA

  • Tube

    A = π(D² − d²)4

  • Bars in series

    ΔL = F Σ LiAiEi

Assumptions
  • The force acts along the bar's axis through the centroid, so stress is uniform across the section.
  • Stress is uniform only away from the ends, holes and changes of section (Saint-Venant's principle). Local stresses there are higher.
  • The material stays linear-elastic, below its yield strength.
  • In compression the same formulas hold for short bars. Slender bars buckle first.
  • Material properties are typical values. Use the supplier's figures for design.

Derivation
  1. Stress is force per unit area: σ = F/A.
  2. Strain is extension per unit length: ε = ΔL/L.
  3. In the elastic range they are proportional (Hooke's law): σ = Eε.
  4. Combine them: ΔL = εL = σL/E = FL/AE.