Structural Steel Design / Single-storey buildings and portal frames

Collapse mechanisms, and the haunch

A portal is one of the few structures designed directly by plastic collapse. Three mechanisms, three work equations, and the largest answer wins.

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.

Corrections checked in this pass

Unequal hinge capacities enter each work equation. 826 kNm weaker eaves/rafter capacity gives candidate factor about 1.396 in place of 2.18 from the column alone. Candidates explicitly upper bounds; haunch interpolation is signed and conditional.

Checks needed for project use

A steel-design reviewer must check the first-generation calculation profiles, second-order analysis assumptions, connection rules and execution references against the applicable standards and amendments.
Optimised hinge locations, pitched geometry, actual haunch moment redistribution, global/member buckling, hinge restraint and rotation capacity. Passing finite candidates cannot establish adequacy.

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

A portal frame is the most-built structural form in the country and it is designed in a way nothing else in this course has been: by working out how it would collapse, and making it strong enough that it does not. That is plastic design, and for a portal it is short enough to do by hand — three mechanisms, three work equations, and the upper-bound theorem to tell you which answer to believe. The haunch then makes the result economic, and the reason it is the length it is falls straight out of the bending moment diagram.

By the end of this lesson you should be able to

Derive each of the three collapse mechanisms
Apply the upper-bound theorem to choose between them
Say what decides which mechanism governs
Quantify what a haunch saves

What you should already know

Plastic hinges and the plastic moment (Module 6)
Virtual work and collapse mechanisms — the Structural Analysis Fundamentals course covers this
Load combinations, including the favourable-permanent case (Module 2)
Base fixity and what it is worth (Module 14)

From first principles

The three collapse mechanisms of a portal frame

We want to show: Derive the plastic moment required by each of the three mechanisms of a pinned-base portal, and see which one governs and why.

A frame collapses when enough plastic hinges have formed to turn it into a mechanism — something that can move without any further increase in load. There are only a few ways a portal can do that: it can sag, it can lean, or it can do both at once. Each way is a mechanism, and each has a work equation. The frame has to be strong enough for the worst of them, and the upper-bound theorem says the worst is the one demanding the LARGEST plastic moment.

Try it

Three mechanisms, and which one governs

Three candidate mechanisms for a pinned rectangular frame with uniform plastic capacity. The solid bar identifies the largest required Mp within this set; a real pitched or haunched frame needs its own analysis.

Span30 m

Eaves height6.0 m

Vertical load9.45 kN/m

Horizontal load40 kN

Haunch length10 % of span

Beam mechanism: 532 kNm
Sway mechanism: 120 kNm
Combined mechanism: 592 kNm
Governing mechanism: COMBINED
Plastic moment required: 592 kNm
Load ratio wL² over Hh: 35.4
Crossover load w: 1.07 kN/m
Haunch length: 3.0 m
Assumed-diagram moment beyond haunch: 420 kNm
Change within the assumed linear diagram: 29 %

The combined candidate requires 591.56 kNm for this uniform-capacity, pinned rectangular model. The other candidate work equations remain visible. This is not a verified collapse load for a pitched or haunched frame. At 3.00 m on the assumed linear signed moment diagram, M = -420.01 kNm. The remaining interval has maximum absolute moment 420.01 kNm. This interpolation does not model the stiffness redistribution caused by adding a haunch.

Things worth trying

The default equal-capacity rectangular model gives combined required Mp =591.56 kNm.
With no horizontal action, beam and combined candidates coincide; with no vertical action, sway governs this candidate set.
The haunch readout locates a point on an assumed linear moment diagram. It does not recalculate frame actions after stiffness changes.
For unequal member or joint capacities, write each hinge’s work separately and search relevant hinge positions.

Worked example

A 30 m portal frame

Given

30 m span, 6 m to eaves, 6° pitch, frames at 6 m centres, pinned bases
Roof: permanent 0.5 kPa, variable 0.6 kPa. At ULS 1.35 × 0.5 + 1.5 × 0.6 = 1.575 kPa
So w = 1.575 × 6 = 9.45 kN/m on each frame
Horizontal load at eaves, including the equivalent horizontal force: H = 40 kN
For a separate un-haunched rectangular mechanism model, assume eaves hinge capacities Me = 826 kNm and mid-rafter capacity Mm = 826 kNm. The nominal column capacity 1289 kNm does not control a hinge that can form in the weaker adjoining rafter.

Find

Calculate candidate first-order mechanism multipliers for the explicitly assumed hinge capacities; do not infer frame adequacy.

Assumptions

Uniform plastic moment for the mechanism calculation — the haunch is dealt with separately
A low-pitch frame treated as flat for the mechanisms
Class 1 sections throughout, so the plastic modulus is available

The haunch

A haunch adds resistance and stiffness near the eaves and facilitates the joint. Its length is often around 10% of the span, but the complete frame analysis determines the actions.

For a separate illustrative linear moment diagram from −592 kNm at the eaves to +266 kNm at the apex, a 3 m haunch on a 30 m span ends at 20% of the half-span. Its signed moment is −592 + 0.2(266 + 592) = −420.4 kNm. The un-haunched region of this assumed diagram has a maximum magnitude of 420.4 kNm, a 29% reduction from the eaves magnitude.

That calculation locates a point on an assumed diagram. It does not calculate the redistribution caused by adding a haunch. A UDL produces a curved moment diagram, so the linear interpolation must not be presented as its exact solution.

Practice

A portal spans 30 m and carries w = 9.45 kN/m on the rafter. What plastic moment does the BEAM mechanism require, in kNm?

Practice

The same frame is 6 m to eaves with H = 40 kN, pinned bases. What does the SWAY mechanism require, in kNm?

Practice

What does the COMBINED mechanism require for that frame, in kNm?

Practice

That frame's column has M_c,Rd = 1289 kNm. What is the collapse load factor λ?

Check yourself

Why does the combined mechanism demand more plastic moment than the beam mechanism when H is positive?

Summary

Beam mechanism: M_p = wL²/16, from a three-hinge rafter mechanism
Sway mechanism: M_p = Hh/2, with pinned bases and no hinge at the feet
Combined: M_p = wL²/16 + Hh/4, because one hinge cancels but the work does not
The upper-bound theorem says design for the LARGEST
The combined governs whenever wL² > 4Hh, which is nearly always
The 30 m frame needed 592 kNm and the column gives λ = 2.18
λ is a margin above FACTORED load, not a factor of safety
A 10% haunch cut the rafter's design moment from 592 to 420 kNm