Skip to main content

Idealized stress-strain diagram of steel

Short answer

Mild steel is idealized as elastic-perfectly plastic with a yield plateau at f_y, elastic modulus E=2×10^5 N/mm², and yield strain 0.00125 for f_y=250. High tensile steel has no distinct yield point; its yield is the 0.2% proof stress, idealized as elastic up to that stress then flat, with lower ductility.

Show the idealized stress-strain diagram for mild steel and high tensile steel used in the design of steel structures.

The real stress-strain curve is idealised for design (IS 800:2007).

Mild steel (E250)

 stress
  fu |            ___------.
     |         _-/          \
  fy |-------+==============+   <- yield plateau
     |      /|
     |     / |
     |    /  |  E = 2x10^5
     |   /   |
     |__/____|______________|____ strain
        0.00125   ~0.014    ~0.2
  • Elastic line up to fyf_y with slope E=2×105E=2\times10^5 N/mm²; yield strain εy=fy/E=0.00125\varepsilon_y=f_y/E=0.00125 for fy=250f_y=250.
  • Constant stress fyf_y (yield plateau) up to roughly 10-15 times εy\varepsilon_y (strain hardening starts at about 0.014).
  • Strain hardening up to the ultimate stress fuf_u (≈410\approx410 N/mm²) at a strain of about 0.15-0.2, then necking and fracture.
  • For design the curve is idealised as elastic-perfectly plastic (bilinear): σ=Eε\sigma = E\varepsilon up to fyf_y, then σ=fy\sigma=f_y.

High tensile steel (E350-E450, cold-worked)

 stress
  fu |           ___---
     |        _-/
 0.2%|------_/
 proof      /
     |     /
     |    /
     |___/____|__________ strain
       0.002 offset
  • No definite yield point or plateau; the stress rises smoothly.
  • The yield stress is taken as the 0.2% proof stress (found by a line parallel to the elastic part, drawn from 0.002 strain).
  • Idealised as elastic up to the proof stress and then flat (or rising slightly). Ductility is lower than that of mild steel.

✓ Answer checked by Claude Opus 5.5 (high effort)

All Design of Steel and Timber Structure chapter 1 questions →