Module 2 · Lesson 2.3

Combinations of actions

Building the ULS and SLS combinations for a real floor, and why the largest variable action is not always the leading one.

Why this matters

A structure is rarely loaded by one thing. It has weight, people, wind, perhaps snow, and the design must decide what to assume about them happening together. Assuming they all peak simultaneously is absurdly conservative; assuming they never do is unsafe. The combination rules are the compromise, and once you see what the ψ factors are for, the rules stop being arbitrary bookkeeping and become obvious.

By the end of this lesson you should be able to

Write the ULS fundamental combination and explain each term
Explain what ψ₀, ψ₁ and ψ₂ each represent
Build the three serviceability combinations and say what each is used for
Identify the leading variable action by trying each in turn
Assemble the full set of load cases for a simple floor

What you should already know

Classification of actions (previous lesson)
Partial factors and what they cover (this module, lesson 1)

The problem the ψ factors solve

Consider an office floor carrying an imposed load and a building carrying wind. The imposed load reaches its characteristic value perhaps once in the building's life — a party, a move, a filing room. The wind reaches its characteristic value perhaps once in fifty years. The chance of both happening in the same hour is vanishingly small.

If you designed for both at full value simultaneously, you would be designing for an event far rarer than the target reliability requires, and the structure would be uneconomic for no gain in safety.

The resolution is simple in principle. One variable action is allowed to be at its full characteristic value — the leading action. Every other variable action is reduced, because it is being asked to be unusually large at exactly the moment another action is also unusually large.

ULS fundamental combination

E_{d} = j \sum γ_{G, j} G_{k, j} + γ_{Q, 1} Q_{k, 1} + i > 1 \sum γ_{Q, i} ψ_{0, i} Q_{k, i}

What it calculates: The design effect for persistent and transient design situations — the everyday ULS check.

$G_{k, j}$: Characteristic value of permanent action j
$Q_{k, 1}$: Characteristic value of the LEADING variable action
$Q_{k, i}$: Characteristic value of an accompanying variable action
$ψ_{0, i}$: Combination factor — reduces an accompanying action (—)
$γ_{G, j}$: 1.35 if unfavourable, 1.0 if favourable
$γ_{Q, i}$: 1.5 if unfavourable, 0 if favourable

This assumes

Every variable action is tried in turn as the leading one
The γ and ψ values are nationally determined and require verification

In plain terms: Read it as: all the permanent load, plus one variable action at full value, plus everything else at a discount. The discount is the ψ₀ factor, and it exists purely because simultaneity is unlikely.

The three ψ factors

They are three different reductions answering three different questions about the same variable action.

ψ₀ — the combination value. 'If some other action is at its extreme, how large is this one likely to be at the same moment?' Used in ULS combinations and the characteristic SLS combination.

ψ₁ — the frequent value. 'What value is exceeded only a small fraction of the time?' Used for reversible serviceability effects and in accidental situations.

ψ₂ — the quasi-permanent value. 'What part of this action is essentially always there?' For an office floor, the desks, chairs, files and equipment never leave; only the crowd of people is transient. Used for long-term effects — creep, long-term deflection, crack width.

The ordering is always ψ₀ > ψ₁ > ψ₂, and each has a clear physical meaning. ψ₂ in particular is not a safety factor at all: it is a statement about what fraction of the imposed load is furniture.

The serviceability combinations

Characteristic — G + Q₁ + Σψ₀Qᵢ. The most severe SLS case, for irreversible effects: cracking of a finish that will not heal, permanent damage.

Frequent — G + ψ₁Q₁ + Σψ₂Qᵢ. For reversible effects and for checking that a condition is not exceeded too often.

Quasi-permanent — G + Σψ₂Qᵢ. The long-term average. This is the one to use for deflection and crack width, because those are governed by what the structure carries for years, not for minutes.

All three use unfactored loads. None of them has a γ in it.

Worked example

A complete set of combinations for an office floor

Given

200 mm reinforced concrete slab, density 25 kN/m³
Screed, finishes and services: 1.5 kN/m² (permanent)
Fixed partitions: 1.0 kN/m² (permanent)
Office imposed load: 2.5 kN/m², with ψ₀ = 0.7, ψ₁ = 0.5, ψ₂ = 0.3
No wind on this internal floor

Find

The ULS design load and the three serviceability loads, per square metre.

Assumptions

One variable action only, so it is automatically the leading one
All actions unfavourable for the sagging check being designed
The ψ values are nationally determined and require verification

Try it

Build the load combinations

Change the actions and the ψ factors, and switch which variable action leads. The governing case is not always the one with the biggest action.

Permanent g_k20 kN/m

Imposed q_k10 kN/m

Wind w_k8 kN/m

ψ₀ imposed0.70

ψ₀ wind0.50

ψ₂ imposed0.30

Leading variable action

Terms in the ultimate limit state combination
Action	Characteristic	Factor	Design
Permanent	20.0	1.350	27.00
Imposed	10.0	1.500	15.00
Wind	8.0	0.750	6.00

ULS, Imposed leading: 48.00 kN/m
ULS, other case: 49.50 kN/m
SLS characteristic: 34.00 kN/m
SLS quasi-permanent: 23.00 kN/m
Quasi-permanent / ULS: 46%

The OTHER case governs, at 49.50 kN/m against 48.00. Both must be checked.

Why each factor was chosen

Permanent: Permanent, unfavourable — factored up.
Imposed: Leading variable action — full factored value.
Wind: Accompanying variable action — reduced by ψ₀ = 0.5, since it is unlikely to peak at the same moment as the leading action.

Things worth trying

Set the wind smaller than the imposed load, then lower its ψ₀. The wind-leading case can still govern — size is not the only thing that matters.
Make ψ₀ equal for both. Now the larger action always leads, and the surprise disappears.
Watch the quasi-permanent line. It is what the deflection check uses. On a floor with no wind it is typically 50 to 65% of the ULS load; with wind in the combination it falls further, because wind adds to the ULS total and contributes nothing at all to the quasi-permanent one.
Raise ψ₂ towards 1. That is a warehouse rather than an office: the stored goods never leave, so almost all the imposed load is permanent for deflection.

Worked example

Two variable actions — and the leading one is not the bigger one

Given

Permanent action on a frame: g_k = 20 kN/m
Imposed: q_k = 10 kN/m, ψ₀ = 0.7
Wind: w_k = 8 kN/m, ψ₀ = 0.5
Both variable actions unfavourable for this check

Find

The governing ULS combination.

Assumptions

Both variable actions act in the same sense for the effect being checked
Each must be tried as the leading action — there is no shortcut

Practice

A 250 mm slab (25 kN/m³) carries finishes of 1.2 kN/m² and an imposed load of 4.0 kN/m². What is the ULS design load, in kN/m², using 1.35 and 1.5?

Practice

For the same slab, what is the quasi-permanent load used for the deflection check, in kN/m²? Take ψ₂ = 0.3.

Practice

A structure carries g_k = 15 kN/m, an imposed load of 12 kN/m (ψ₀ = 0.7) and snow of 6 kN/m (ψ₀ = 0.5). What is the governing ULS load, in kN/m?

Summary

One variable action leads at full value; the rest are reduced by ψ₀
ψ₀ combination, ψ₁ frequent, ψ₂ quasi-permanent — always in that decreasing order
ψ₂ is not a safety factor: it is the fraction of the imposed load that is furniture
Deflection and crack width use the quasi-permanent combination
Try EVERY variable action as the leading one — the biggest does not always win
A typical office floor: about 13.9 kN/m² at ULS, 8.25 kN/m² quasi-permanent

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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