The Chartership Forum

Structural Analysis Fundamentals / Bending of beams

Plastic bending and the plastic hinge

What a ductile section can still do after its extreme fibre has yielded.

Lesson review: expert verification remains open

This lesson has received a full-context editorial review and a check of its technical explanations. Selected arithmetic has been checked separately. This is not independent engineering sign-off.

Checks needed for project use

  • An independent structural reviewer must check remaining derivations, diagrams, sign conventions and the interactive models across their full input ranges.
  • Independent review of diagrams, idealisations and full interactive input ranges remains required.

These are educational models and exercises. Check their assumptions before using the results. The full technical content and every diagram have not yet been independently verified.

Concrete and steel calculations retain their stated first-generation Eurocode teaching basis. Use the standards and amendments specified for your project; do not mix generations without an appropriate design basis.

Read the lesson register and outstanding review notes

Why this matters

First yield marks the limit of an entirely elastic section calculation. With suitable ductility, restraint and section proportions, further bending can spread yielding across the section and increase its moment resistance.

By the end of this lesson you should be able to

  • Describe how the stress block develops from elastic to fully plastic
  • Calculate the plastic section modulus and the plastic moment
  • Use the shape factor to compare plastic and yield moments
  • Explain what a plastic hinge is and why it allows redistribution

What you should already know

  • Bending stress σ = My/I and section modulus Z (this module)
  • The stress–strain curve and the yield plateau (Module 8)

For a homogeneous elastic-perfectly-plastic material in pure bending, stress is initially linear. The extreme fibre first reaches yield at My = σy Z.

Further curvature spreads yielded zones inward while an elastic core remains. In the fully plastic limit the stress is +σy on one side and −σy on the other, giving Mp. This idealisation neglects strain hardening and depends on sufficient section and member stability. First yield is not automatically structural failure.

Fully plastic moment
Mp=σyZp,Zp=∑AiyˉiM_p = \sigma_y Z_p, \qquad Z_p = \sum A_i \bar{y}_i

What it calculates: the moment at which the whole section has yielded

Mp
fully plastic moment (N·mm)
Zp
plastic section modulus (mm³)
Ai
area of each half of the section (mm²)
ȳ_i
distance from the plastic neutral axis to that half's centroid (mm)

This assumes

  • The material has a well-defined yield plateau and enough ductility to reach it
  • Rigid-plastic idealisation: strain hardening is ignored, which is conservative
  • The section is stocky enough not to buckle locally before it yields fully

In plain terms: Zp is the first moment of the two areas about the equal-area axis — the plastic counterpart of Z = I/y. For a rectangle Zp = bh²/4, against Z = bh²/6.

The ratio of the two moments is the shape factor:

shape factor = Mp/My = Zp/Z

It depends only on the shape of the section, not on the material or the size. Two values worth carrying:

  • Rectangle: Zp/Z = (bh²/4)/(bh²/6) = 1.5. Half as much again beyond first yield.
  • Universal beam: typically 1.10 to 1.20. Much less, because most of an I-section's material is already in the flanges at nearly the extreme fibre stress, so there is little left to catch up.

That contrast is worth pausing on. The I-section is far more efficient elastically — which is exactly why it has less in reserve once yielding starts.

Predict first

Which has the larger shape factor: a solid rectangular section, or a universal beam of the same depth?

An ideal plastic hinge transmits Mp while accommodating rotation. Real hinge rotation is finite and depends on local and member stability, material ductility, shear, axial force and connection behaviour.

An indeterminate beam may redistribute moment after a hinge forms. For the ideal single-span arrangements here, a propped cantilever mechanism needs two plastic hinges and a fixed-ended beam mechanism needs three. Other failure modes can occur before that mechanism forms.

Worked example

Yield moment, plastic moment and shape factor

Given

  • Rectangular steel section: b = 150 mm, h = 300 mm
  • Yield stress σy = 355 N/mm²

Find

The elastic and plastic section moduli, both moments, and the shape factor.

    Practice

    A rectangular steel section is 150 mm wide and 300 mm deep, with σy = 355 N/mm². What is its plastic section modulus, in mm³?

    Practice

    For that same section, what is the fully plastic moment, in kNm?

    Practice

    A section has an elastic modulus Z = 1.20 × 10⁶ mm³ and a plastic modulus Zp = 1.38 × 10⁶ mm³. What is its shape factor?

    Summary

    • Beyond first yield a ductile section carries more, as the yielded zone spreads inwards
    • Fully plastic: two rectangular stress blocks at ±σy
    • The plastic neutral axis is the equal-area axis, not the centroid
    • Mp = σy Zp; for a rectangle Zp = bh²/4
    • Shape factor Zp/Z is 1.5 for a rectangle, about 1.15 for a universal beam
    • A plastic hinge rotates at constant Mp and allows redistribution — but only if the material is ductile