Module 9 · Lesson 9.2

Frame stability, and which moments get amplified

The same effect at the scale of a whole frame. It turns into a question about lateral stiffness — and about which moments the amplifier is allowed to touch.

Why this matters

A member bows and its axial force acts through the bow. A frame leans and its whole vertical load acts through the lean. It is the same mechanism, and at frame scale it is measured by a single number that can be read off a first-order analysis: α_cr. What makes it worth a lesson of its own is that the remedy is different. A member that fails on amplification usually needs restraint or a bigger section. A frame that fails on α_cr needs stiffness, and no amount of member design supplies it.

By the end of this lesson you should be able to

Calculate α_cr for a storey from its drift under horizontal load
Explain α_cr as a comparison of two stiffnesses
Say when second-order analysis is required and why the threshold differs for plastic analysis
Apply the amplifier to the sway moments and not to the gravity moments
Convert an α_cr shortfall into the stiffness change it actually demands

What you should already know

The magnification series and α_cr (Module 4)
Equivalent horizontal forces from sway imperfections (Module 4)
The amplification factor from the previous lesson

α_cr is a ratio of two stiffnesses

For a storey, the critical multiplier can be written straight from a first-order analysis:

α_cr = (H_Ed / V_Ed) · (h / δ_H)

where H_Ed is the horizontal force on the storey, V_Ed the vertical force, h the storey height and δ_H the relative sway of the storey under H_Ed.

Grouped differently, it reads much better:

α_cr = (H_Ed / δ_H) ÷ (V_Ed / h)

H/δ is the storey's lateral stiffness — force per unit sway. That is what resists the lean.

V/h is the rate at which the vertical load converts sway into overturning — a destabilising stiffness, with the same units. That is what causes it.

So α_cr is simply the ratio of the stiffness you have to the destabilising effect you must resist. At α_cr = 1 they are equal and the storey has nothing left. At α_cr = 10 the lateral stiffness is ten times what the vertical load is consuming, and the amplification is 1/(1 − 0.1) = 1.11.

This reading is worth more than the formula, because it tells you what to change. α_cr is proportional to lateral stiffness and inversely proportional to vertical load. If a storey is too flexible, you need stiffness — bracing, a stiffer frame, a core — and adding steel to columns barely moves it.

Note

'Non-sway' does not mean the frame does not sway

Every frame sways. The classification is about whether the sway is small enough that the extra moments it produces can be ignored.

Above the threshold the frame is treated as non-sway and first-order moments are used directly. Below it, the second-order effects must be accounted for — either by a proper second-order analysis, or by amplifying the first-order sway moments.

The threshold is code-calibrated and nationally determined, and it is different for elastic and plastic analysis: plastic redistribution softens the frame beyond what an elastic α_cr predicts, so the plastic threshold is set higher to compensate. Both values in this course are held in the standards profile and marked as requiring verification.

The threshold is a decision about acceptable error, not a physical boundary. Nothing changes about the frame as it crosses it.

Try it

Drift, α_cr, and where the moments go

A 4 m storey under 180 kN of horizontal force. Change the drift and the vertical load and watch α_cr move. Then change how much of the moment is sway moment, and see what the two common shortcuts cost.

Storey drift22 mm

Vertical load4,200 kN

Sway share of the moment31 %

Global analysis

Storey drift δ: 22 mm
Critical multiplier α_cr: 7.79
Classification: SWAY
Sway amplifier: 1.147
Second-order analysis required: YES
Design moment, sway only: 319 kNm
If everything amplified: 350 kNm
If nothing amplified: 305 kNm
Cost of over-amplifying: 9.7 %
Error from not amplifying: 4.4 %
Stiffness needed for the threshold: 1.28 ×

α_cr = 7.79: first-order sway moments must be amplified by 1.15, or a second-order analysis used. The storey drift of 22.0 mm is what drives this — stiffness, not strength, is the governing quantity.

Things worth trying

Start at 22 mm. α_cr = 7.79, the storey is a sway frame, and the three bars are 305, 319 and 350 kNm — the two shortcuts bracket the right answer on either side.
Take the drift down to 12 mm. α_cr crosses 10 and the classification flips to non-sway. Nothing about the frame has changed except its stiffness — the threshold is a decision about acceptable error, not a physical boundary.
Now switch to plastic analysis at the same drift. Nothing changes — and that is worth knowing. The Eurocode RECOMMENDS a higher threshold for plastic analysis (15 against 10), on the reasoning that plastic redistribution softens the frame beyond what the elastic α_cr predicts. The UK National Annex, NA.2.9, replaces it with 10 for clad structures, so in the UK the method does not change the classification.
Where the method DOES still change it is a portal frame under gravity loads only, which NA.2.9 relaxes all the way to 5 — subject to limits on span and rise. That relaxation carries conditions, so this lab does not apply it automatically.
Push the drift to 60 mm at 8000 kN. α_cr falls towards 1 and the storey approaches instability. No member design fixes that — the storey needs stiffness.
Set the sway share to 100%. The 'sway only' and 'amplify everything' bars converge, because there is nothing else to amplify wrongly.
Now set it to 10%. Amplifying everything becomes expensive and amplifying nothing becomes almost harmless — which is exactly why the two errors are so often made in the wrong situations.
Watch the stiffness figure as you move the drift. It is the most useful number here, because it converts a failed check into the design change that would fix it.

Worked example

A storey that is too flexible

Given

A storey of a moment-resisting frame, 4.0 m high
Total horizontal force at the storey H_Ed = 180 kN, including the equivalent horizontal force from the sway imperfection
Total vertical force at the storey V_Ed = 4200 kN
First-order relative sway under H_Ed: δ_H = 22 mm
Column design moments: 210 kNm from the gravity (non-sway) case, 95 kNm from the sway case
Elastic global analysis

Find

Whether second-order effects must be included, and the design moment if so.

Assumptions

The α_cr threshold is calibrated and nationally determined — held in the profile, unverified here
The amplified sway method assumes the sway deflection has essentially the shape of the sway buckling mode

Member and frame effects are different problems

Both are second-order effects and both use the same 1/(1 − ratio) form, which makes them easy to conflate. They behave completely differently in design.

	Member (P–δ)	Frame (P–Δ)
What deflects	The member, between its ends	The whole storey, sideways
Driving ratio	N_Ed/N_cr for the member	1/α_cr for the storey
Made worse by	A slender member	A flexible frame
Fixed by	Restraint, or a stiffer member	Bracing, or a stiffer frame
Grade helps?	No	No
Visible in a first-order analysis?	No	No, but α_cr can be read from one

The last row is the practical one. Neither effect appears in the output of a linear analysis, but the frame effect can be diagnosed from one in a single line of arithmetic, which is why α_cr is worth computing early on every project. There is no equally cheap test for the member effect — it has to be checked member by member.

And they must not be counted twice. If the analysis is genuinely second-order and includes the sway, the frame effect is already in the results and must not be amplified again. Module 4's allocation table is the discipline that keeps this straight.

Practice

A storey carries V_Ed = 4200 kN, is 4.0 m high, and sways 22 mm under a horizontal force of 180 kN. What is α_cr?

Practice

What is the sway amplification factor at α_cr = 7.79?

Practice

Gravity moments are 210 kNm and sway moments 95 kNm. With an amplifier of 1.147 applied correctly, what is the design moment in kNm?

Practice

A storey has α_cr = 7.79 and drifts 22 mm. What drift would be needed to reach α_cr = 10, in mm?

Check yourself

A storey has α_cr = 6. The engineer increases every column size by one serial size and re-runs the analysis. What is the most likely outcome?

Summary

α_cr = (H/V)(h/δ) — the storey's lateral stiffness against the destabilising rate
It can be read from a first-order analysis in one line
α_cr = 7.79 in the worked example: a sway frame, amplifier 1.147
Amplify the SWAY moments only: 210 + 1.147 × 95 = 319 kNm
Amplifying everything gave 350 kNm (9.7% wasteful); amplifying nothing gave 305 (4.4% unsafe)
Reaching α_cr = 10 needed 1.28 times the lateral stiffness — a bracing decision
The threshold differs for plastic analysis because redistribution softens the frame
Member and frame second-order effects are the same algebra and different design problems

Progress is kept in this browser only.

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

αcr is a ratio of two stiffnesses

Drift, αcr, and where the moments go

A storey that is too flexible

Member and frame effects are different problems

α_cr is a ratio of two stiffnesses

Drift, α_cr, and where the moments go