Engineer's Notebook §1.1

Hooke's law

Up to its elastic limit, a material's stress is proportional to its strain. The constant of proportionality is Young's modulus:

σ = Eε
σ
stress
Pa
E
Young's modulus
Pa
ε
strain
—
fy
yield strength
Pa
µε
microstrain, 10⁻⁶
—
τ
shear stress
Pa
G
shear modulus
Pa
γ
shear strain
rad
F
force
N
x
extension
m
k
stiffness
N/m
A stress–strain graph. A straight elastic line rises to the yield strength, then an idealised flat line continues. A point marks the chosen strain. STRAIN ε, µεSTRESS σ, MPa 1,000 σ = 200 MPa 275 10,000
Fig. 1.1-1Beyond yield the dashed line is an idealised plateau; real metals harden.
εy = fy/E1.38 × 10⁻³
ε1 × 10⁻³
σ = Eε200 MPa

Worked example

Uses the values in Fig. 1.1-1. Change them and the working updates.

Given
  • Esteel S275200 GPa
  • εstrain1,000 µε
Find

The stress σ.

Solution
ε = 1,000 × 10⁻⁶ = 1 × 10⁻³ σ = Eε = 2 × 10⁵ MPa × 1 × 10⁻³ = 200 MPa
Answer
σ = 200 MPa

That is 72.7% of the yield strength of steel S275 (275 MPa).

Related forms

  • Shear

    τ = Gγ

  • Spring

    F = kx

  • Bar as a spring

    k = AEL

Assumptions
  • The material is linear-elastic: below its limit of proportionality, which for most metals is close to yield.
  • The material is isotropic and the stress is uniaxial.
  • Beyond yield the demo draws an idealised flat plateau (elastic–perfectly plastic). Real metals strain-harden.
  • Material properties are typical values. Use the supplier's figures for design.

Derivation
  1. In a tensile test, stress is σ = F/A and strain is ε = ΔL/L.
  2. Plotted against each other, the first part of the curve is a straight line through the origin.
  3. Its slope is Young's modulus: E = σ/ε, so σ = Eε.

Robert Hooke published it for springs in 1678 as ut tensio, sic vis: as the extension, so the force.