Module 8 · Lesson 8.1
Trusses and frames
What the pin-jointed idealisation omits, and where a connection stops being pinned.
Why this matters
Two idealisations carry most of structural steelwork: that a truss is pin-jointed, and that a frame connection is either rigid or pinned. Both are extremely useful and neither is true.
The interesting question is not whether they are true but how far wrong they are, and in which direction — which is answerable by building the model both ways.
By the end of this lesson you should be able to
- Quantify what the pin-jointed idealisation omits
- Explain why a heavier chord attracts more secondary bending
- Sweep a connection stiffness and find where the moment actually changes
- Use the free bending moment as a check on any portal result
What you should already know
- Module 7's releases — a truss is a frame with every member released
- Module 5's Warren truss, which is the same model used again here
The pin-jointed truss
Every textbook truss has frictionless pins at every joint. Every real truss has chords that run continuously through the panel points, welded or bolted, and are bent by the joint rotations.
Model the same 24 m Warren truss both ways — pin-jointed bars, then continuous chords with pin-ended web members — and the comparison is reassuring in one respect and instructive in another:
| Pin-jointed | Continuous chords | |
|---|---|---|
| Largest bottom chord force | 340.0 kN | 339.1 kN |
| Bending in that chord | 0 | 1.41 kN·m |
| Midspan deflection | 20.30 mm | 20.25 mm |
The axial force is within 0.3 % and the deflection within 0.3 %. The idealisation is excellent for the things it was meant for.
What it omits is the bending, and 1.41 kN·m in a member sized for 340 kN axial is not nothing: through the section it accounts for about 7.7 % of the peak stress. A member checked at 95 % utilisation on axial force alone is over.
The counter-intuitive part
Make the chord stiffer and the secondary bending goes up, not down.
Bending follows stiffness. A continuous member picks up moment in proportion to how strongly it resists the rotation being imposed on it, so a heavy chord attracts more of it than a light one. A truss cannot be made safe against secondary bending by making the chords bigger; it can only be made safe by detailing the joints so the rotation is not imposed, or by checking the combined stress.
Where a connection stops being pinned
Module 7 established that a release is a spring of zero. Sweeping that spring on a 12 m portal with fixed bases at 15 kN/m:
| Connection stiffness | Eaves moment | Midspan moment | Share of rigid |
|---|---|---|---|
| Pinned (0) | 0 | 270.0 | 0 % |
| 1 000 kN·m/rad | 38.2 | 231.8 | 26 % |
| 3 000 | 75.8 | 194.2 | 51 % |
| 10 000 | 115.5 | 154.5 | 78 % |
| 30 000 | 135.8 | 134.2 | 91 % |
| Rigid (∞) | 148.9 | 121.1 | 100 % |
Two things are worth taking from that table.
The eaves and midspan moments always add to 270 kN·m, which is wL²/8 — the free bending moment of the span. That is not a property of this frame; it is statics, and it holds for every row. It is therefore the check to run on any portal result before reading anything else.
Half the rigid moment arrives at about 1.65 EI/L. The beam's EI/L here is 1 750 kN·m/rad, and the 50 % point is at 2 896. Ninety per cent needs 14.9 EI/L — nine times as much stiffness for the last 40 %.
So the curve is steep exactly where real connections are, and flat well above them. A connection an engineer would describe as 'quite stiff' is often at 50–70 % of the rigid moment, and calling it rigid over-predicts the eaves moment and under-predicts the midspan by a similar margin — in opposite directions, on the same member.
The linkage
One combination is not a frame at all. Pinned bases and pinned connections give four pins in a four-bar arrangement, which is a mechanism: nothing resists sway. A solver should refuse it, and it is worth knowing that some will instead return a very large number without comment, because the mechanism's pivot survives a naive positive-definiteness test.
What it calculates: The check that any portal result must satisfy, whatever is assumed about the connections
- w
- uniformly distributed load on the beam (kN/m)
- L
- span (m)
- M
- moment (kN·m)
This assumes
- A uniformly loaded beam between two supports, however those supports are restrained
In plain terms: This is statics rather than a property of any particular frame, so it holds at every point on the connection sweep. If a model's two moments do not add to wL²/8, the problem is the model rather than the connection assumption.
Try it
Truss modelling workshop
A 24 m Warren truss as pin-jointed bars, and again as a frame with continuous chords.
Bending follows stiffness — try increasing it.
- Chord force, pin-jointed
- 340.0 kN
- Chord force, continuous chords
- 339.1 kN
- Secondary bending in that chord
- 1.41 kN·m
- Bending as a share of peak stress
- 7.7 %
- Midspan deflection, pin-jointed
- 20.30 mm
- Midspan deflection, continuous
- 20.25 mm
not present in the pin-jointed model at all
And the beam analogy, for the same truss
- Midspan moment, actual point loads
- 1080 kN·m
- Midspan moment, loads smeared to a UDL
- 900 kN·m
- Error from smearing
- 16.7 %
- Chord force ≈ M/d
- 360 kN
What this shows: The idealisation gets the axial force right and omits the bending entirely — and a stiffer chord attracts more of it, not less.
Try it
Frame connection explorer
Sweep the beam-to-column connection from a pin to full rigidity. Their sum is the free bending moment at every point.
The beam's own EI/L is 1750 kN·m/rad — this is 1.71 × that.
- Eaves moment
- 75.78 kN·m
- Midspan moment
- 194.22 kN·m
- Sum
- 270.00 kN·m
- Share of the rigid eaves moment
- 50.9 %
- Sway at eaves
- 0.023 mm
- Rigid connection would give
- 148.93 kN·m
- Pinned connection would give
- 270.00 kN·m at midspan
wL²/8 = 270.00 — equal at every stiffness
The two moments always sum to the free bending moment. That identity is the fastest check available on any portal result — if a model’s moments do not add to wL²/8, the problem is the model rather than the connection assumption.
What this shows: Rigid and pinned are the ends of one parameter, and the curve is steep exactly where real connections are.
Worked example
A connection assumed rigid that is half-stiff
Given
- A 12 m portal, fixed bases, 15 kN/m on the beam
- The design assumes rigid connections
- The connections as detailed have a rotational stiffness of about 3 000 kN·m/rad
Find
What the design used and what the frame will do
Practice
A portal's eaves moment is 75.8 kN·m and the free bending moment wL²/8 is 270 kN·m. What is the midspan sagging moment?
Practice
A truss chord carries 340 kN axial. The pin-jointed model reports 340 kN and the continuous model 339.1 kN. What is the percentage difference in axial force?
Check yourself
A truss chord is upgraded to a heavier section. What happens to the secondary bending in it?
Worked example
Checking a portal result in ninety seconds
Given
- A model reports an eaves moment of 96 kN·m and a midspan sagging moment of 210 kN·m
- The beam spans 14 m under 12 kN/m
Find
Whether the result can be right
Check yourself
Why does the eaves-plus-midspan identity hold at every connection stiffness?
Check yourself
The pin-jointed truss idealisation gets the axial forces within a fraction of a per cent. What does that justify?
Check yourself
What makes a portal with pinned bases and pinned connections a mechanism?
Summary
- The pin-jointed truss gets the chord force within 0.3 % and the deflection within 0.3 %
- It omits bending entirely — here 7.7 % of the chord's peak stress
- A stiffer chord attracts more secondary bending, not less
- Eaves plus midspan moment equals wL²/8 at every connection stiffness
- Half the rigid moment arrives at about 1.65 EI/L; ninety per cent needs 14.9
- Pinned bases with pinned connections is a linkage, not a frame
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