Module 15 · Lesson 15.1
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.
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
A pinned-base portal. All three mechanisms are shown; the solid bar is the one demanding the most, which is the one the frame must be designed for.
- 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
- Rafter design moment
- 420 kNm
- Saved by the haunch
- 29 %
The COMBINED mechanism governs at 592 kNm, above the beam mechanism's 532 and the sway mechanism's 120. This is the usual case for a real portal: neither pure sagging nor pure sway, but a hinge pattern that combines them, and it always demands more than either alone.
Things worth trying
- Start at the defaults — 30 m, 6 m, 9.45 kN/m, 40 kN. The combined mechanism governs at 592 kNm, and wL²/Hh is 35: the vertical load dominates completely.
- Take the vertical load to zero. The beam bar vanishes, the combined falls to HALF the sway value, and the sway mechanism takes over. That is right — the combined assumes a rafter hinge that has nothing to form it.
- Now bring the vertical load back slowly and find where the sway and combined bars cross. It is at wL² = 4Hh, and the readout gives the crossover load directly.
- Take the horizontal load to zero instead. The combined bar drops exactly onto the beam bar — with no sway there is nothing to combine with.
- Increase the span by 20% and then the height by 20%, separately. The span costs far more, because it enters squared and the height only linearly.
- Now the haunch. Take it from 0 to 20% and watch the rafter's design moment fall — 592 to 420 kNm at 10%, and to 266 at 20%.
- Push the haunch past 20%. The saving stops growing, because the apex moment has taken over as the point that sizes the rafter. That is why haunches are about 10% and not 30%.
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
- Column 610 × 229 × 113 UB in S355; rafter 533 × 210 × 92 UB
Find
Which mechanism governs, and whether the frame is adequate.
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
The bending moment in a portal peaks at the eaves and falls away along the rafter, passing through zero and reversing before the apex. A rafter sized for the eaves moment is therefore enormously over-strong everywhere else.
The haunch fixes that by making the frame deeper only where the moment is largest — usually a length of the same section cut diagonally and welded under the rafter, so it costs a piece of rafter rather than a new section.
For the 30 m frame, with an eaves moment of 592 kNm and an apex moment of 266:
| Haunch length | Moment at the haunch end | Rafter design moment | Saving |
|---|---|---|---|
| 5% of span, 1.5 m | 506 kNm | 506 kNm | 15% |
| 10% of span, 3.0 m | 420 kNm | 420 kNm | 29% |
| 15% of span, 4.5 m | 334 kNm | 334 kNm | 44% |
| 20% of span, 6.0 m | 248 kNm | 266 kNm | 55% |
Notice the last row. Past about 20% the moment at the haunch end has fallen below the apex moment, and the apex takes over as the governing point for the rafter. Beyond that a longer haunch buys nothing.
The usual 10% is an engineering recommendation rather than an optimum. It is roughly where the saving is still growing quickly, the haunch is still short enough to fabricate from one cut, and the plastic hinge — which forms at the end of the haunch, not at the eaves — is still comfortably within the rafter.
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 Mc,Rd = 1289 kNm. What is the collapse load factor λ?
Check yourself
Why does the combined mechanism demand MORE plastic moment than either of the mechanisms it is made from?
Summary
- Beam mechanism: Mp = wL²/16, from a three-hinge rafter mechanism
- Sway mechanism: Mp = Hh/2, with pinned bases and no hinge at the feet
- Combined: Mp = 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
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