Module 9 · Lesson 9.5
Plastic bending and the plastic hinge
What a ductile section can still do after its extreme fibre has yielded.
Why this matters
Elastic theory says a beam has failed the moment its extreme fibre reaches yield. A ductile steel beam has not: it carries substantially more, because the yielded fibres hold their stress while the ones further in catch up. Understanding how much more is the difference between elastic and plastic design, and it is where the idea of a collapse mechanism begins.
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)
Load a ductile beam and follow the stress block through four stages.
Fully elastic. Stress varies linearly, zero at the neutral axis, greatest at the extreme fibres. Everything so far in this module.
First yield. The extreme fibre reaches σy. The moment at this instant is the yield moment, My = σy Z. Elastic theory calls this failure.
Partially plastic. Load further. The outer fibres cannot take more stress — they are on the yield plateau — but they go on straining. Fibres further in are still elastic and pick up more. A yielded zone spreads inwards from both faces, leaving a shrinking elastic core.
Fully plastic. In the limit the elastic core vanishes. The stress block becomes two rectangles: σy in compression over one part of the section and σy in tension over the other. The moment now is the plastic moment Mp.
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?
Once a cross-section reaches Mp it can rotate without taking any more moment. That is a plastic hinge: not a real hinge, but a zone that behaves like one while continuing to transmit Mp.
The consequence is redistribution. In an indeterminate structure, forming a hinge does not cause collapse — it converts the structure into a slightly less indeterminate one, and further load is carried by the remaining structure until enough hinges have formed to make a mechanism. A propped cantilever needs two; a fixed-ended beam needs three.
This is why ductility matters so much. A brittle section cannot form a hinge: it reaches its capacity and breaks, with no rotation and no redistribution. Everything about plastic design rests on the material being able to sustain σy through large strains — the yield plateau from Module 8, doing structural work.
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
This is educational material. It uses simplified examples to teach principles, and must not be relied on for real design or safety-critical decisions. Module overview and checkpoint