Computational Engineering / Modelling structural systems

Trusses and frames

What the pin-jointed idealisation omits, and where a connection stops being pinned.

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Why this matters

Pin-jointed trusses and idealised rigid or pinned frame connections are useful models when their assumptions suit the question. Compare alternative representations to assess the effect of chord continuity and connection stiffness on the response quantities used for design.

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. Some real trusses have continuous chords whose joint rotations introduce bending. Other trusses have different joint and chord arrangements; represent the relevant detail.

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. Changing a chord section changes both stiffness and resistance and may alter joint rotations throughout the truss. Reanalyse the structure and check the combined actions and resistance; increased moment alone does not establish whether the revised section is adequate.

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.

For this symmetric, uniformly loaded horizontal span, the eaves hogging magnitude plus the midspan sagging moment equals wL²/8 = 270 kN·m. With unequal end moments, use their average with consistent signs. Other loading or geometry requires the corresponding equilibrium relation.

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.

The free bending moment identity

M_{\text{eaves}} + M_{\text{midspan}} = \frac{wL^2}{8} \quad \text{for every connection stiffness}

What it calculates: A moment-equilibrium check for the stated horizontal span under uniform transverse loading, with equal end moments

w: uniformly distributed load on the beam (kN/m)
L: span (m)
M: moment (kN·m)

This assumes

A horizontal beam under uniform transverse loading with equal end moments; use the average signed end moment if they differ

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.

Truss depth (m)

Chord second moment of area (×10⁻⁵ m⁴)

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

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.

Connection stiffness (kN·m/rad)

The beam's own EI/L is 1750 kN·m/rad — this is 1.71 × that.

Span (m)

Load (kN/m)

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

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

If chord flexural stiffness increases while its end rotations are held fixed for a local comparison, what happens to the secondary moment?

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