Module 4 · Lesson 4.1

Second-order effects, from first principles

A frame leans, the vertical load acts through the lean, the frame leans further. Where that stops — and where it does not — is the whole of stability analysis.

Why this matters

First-order analysis solves for equilibrium in the undeformed geometry: it works out the forces as if the structure had not moved. Real structures move first and then have to be in equilibrium in their moved position, and for a frame carrying substantial vertical load the difference is not small. Understanding where the amplification comes from — and why it runs away at a particular load — is what makes α_cr something more than a number the software prints.

By the end of this lesson you should be able to

Follow the physical sequence that produces second-order moments
Derive the magnification factor as a geometric series
Interpret α_cr, and say what happens as it approaches 1
Explain why the threshold for neglecting second-order effects is calibrated

What you should already know

Euler buckling and the critical load (Structural Analysis Fundamentals, Module 10)
Structural forms and modelling (Module 3)
Geometric series, at least informally

The physical sequence

Take a frame carrying vertical load P at the top and apply a horizontal load H.

Step one. The frame sways sideways by Δ₁. That is what a first-order analysis reports, and if the vertical load did not exist it would be the whole answer.

Step two. But the vertical load is still there, and the columns are now leaning. So P acts through the offset Δ₁ and produces an additional overturning moment PΔ₁.

Step three. That additional moment causes additional sway Δ₂.

Step four. Which the vertical load also acts through, producing PΔ₂, and so on.

This does not stop after one round. Each increment of sway produces a further overturning moment, which produces a further increment of sway, indefinitely.

The question is not whether the process happens. It is whether it converges.

Where the geometric series comes from

Let the frame's lateral stiffness be K, so the first-order sway is Δ₁ = H/K.

The extra overturning from step two is PΔ₁, which produces extra sway PΔ₁/K. Write r = P/K, so:

Δ₂ = rΔ₁

The same argument applies to every subsequent round, so:

Δ = Δ₁(1 + r + r² + r³ + …)

That is a geometric series, and it sums to Δ₁/(1 − r) — provided r < 1.

Now notice what r is. The frame becomes unstable when the vertical load reaches the value at which the sway no longer converges, which is r = 1, so P_cr = K. Therefore:

r = P/P_cr = 1/α_cr

where α_cr is the factor by which the applied load would have to be multiplied to reach the elastic critical load. And so:

M_II = M_I / (1 − 1/α_cr)

The whole of the standard amplification method is that geometric series.

Where this stops working

What the infinity at α_cr = 1 does and does not mean

As α_cr → 1 the amplification factor → ∞. Two things need saying about that.

What it does mean. In this model the sway increments stop shrinking. Each round of the sequence produces as much extra sway as the last, so the process never settles and there is no equilibrium position. That is a genuine statement about elastic instability.

What it does not mean. No real frame deflects infinitely. Before it could, two things intervene: the material yields, so the stiffness K falls and the whole linear argument stops applying; and the large-displacement geometry ceases to be described by this small-displacement model. The infinity is a property of the model, not a prediction about the building.

The useful reading is the opposite of alarming: the expression tells you how much margin there is. At α_cr = 10 the amplification is 1/(1 − 0.1) = 1.11, about 11%. At α_cr = 3 it is 1.50. At α_cr = 1.5 it is 3.0, and the frame is in serious trouble long before the arithmetic says infinity.

The series also assumes each increment of sway has the same shape as the first. That is why this is an approximation and not a substitute for a proper second-order analysis — though for a regular frame it is a very good one.

Try it

Watch the series converge — or not

Each round of sway produces more overturning, which produces more sway. The bars are the successive increments. While α_cr is comfortably above 1 they shrink fast; as it approaches 1 they stop shrinking and the frame has no equilibrium position.

Critical multiplier α_cr6.0

First-order moment M_I180 kNm

Critical multiplier α_cr: 6.00
Series ratio r = 1/α_cr: 0.167
Amplification 1/(1 − r): 1.200
Sum of the first 8 terms: 1.200
First-order moment M_I: 180 kNm
Second-order moment M_II: 216 kNm
Increase: 20 %
Second order required (elastic)?: YES — below the threshold of 10

α_cr = 6.00 amplifies the first-order moments by 1.20. Second-order effects must be accounted for — either in the analysis, or by amplifying the first-order results.

Things worth trying

Set α_cr to 10 — the threshold. The amplification is 1.11, about 11%, and the bars after the first are already tiny. This is the judgement the threshold encodes: 11% is small enough to absorb.
Drop α_cr to 3. The increments barely shrink and the amplification is 1.50. A frame this flexible is normally stiffened rather than designed for the amplification.
Push α_cr towards 1.1. The bars stop shrinking almost entirely, the 8-term sum falls well short of the closed form, and the method itself is no longer reliable — a warning sign in its own right.
Compare the 8-term sum with the closed form. They agree closely at high α_cr and diverge as it falls, because a slowly converging series needs many more terms. The rate of convergence IS the margin.
Note that the infinity at α_cr = 1 is a property of this model. A real frame yields and sheds stiffness long before then — the expression tells you about margin, not about a predicted deflection.

Worked example

A sway frame, first order and second

Given

Two-storey frame, total vertical design load 4 000 kN
Lateral stiffness such that the elastic critical load is 24 000 kN
First-order design moment at a column base, 180 kNm
Elastic analysis; the frame is unbraced

Find

Whether second-order effects may be neglected, and the design moment if not.

Assumptions

The amplification method is applicable — a regular frame with a dominant sway mode
The α_cr threshold is code-calibrated and is not verified in this course

Predict first

A perfectly symmetric portal frame carries a symmetric vertical load and no horizontal load. A geometrically non-linear (second-order) analysis is run, with no imperfections. What does it report?

Imperfections as equivalent forces

The fix for the problem in that prediction is to give the frame a lean.

A real frame is never plumb. It is erected within a tolerance, and its columns are slightly out of vertical. So the vertical loads do not act exactly through the columns, and they produce overturning even with no wind at all.

Rather than model the lean geometrically, it is replaced by an equivalent horizontal force:

a frame leaning by φ, carrying vertical load V, produces overturning Vφ per unit height — and so does a horizontal force H = φV applied to a plumb frame.

The two are equivalent for overturning, exactly. This is not an approximation; it is a substitution, and it is why imperfections appear in the load cases rather than in the geometry.

Two reductions apply to the basic value, and both are statements about probability rather than mechanics:

Height. Taller frames are relatively straighter, because the erection tolerance does not scale with height.
Number of columns. The more columns in a row, the less likely they all lean the same way.

The imperfection is an equivalent, chosen so the analysis reproduces observed behaviour. It is not the erection tolerance, and it is not a measurement of any real frame.

Practice

A frame has α_cr = 8. By what factor are its first-order sway moments amplified?

Practice

A frame carries 5 000 kN of vertical load and has an elastic critical load of 15 000 kN. What is α_cr?

Practice

A frame level carries 2 400 kN of vertical load and has an initial sway imperfection of 1/250. What equivalent horizontal force should be applied, in kN?

Check yourself

What does a linear buckling analysis (LBA) tell you?

Summary

Sway increments form a geometric series with ratio r = 1/α_cr
M_II = M_I/(1 − 1/α_cr) — the whole amplification method is that sum
The series converges only while α_cr > 1; the infinity is a model property, not a prediction
At α_cr = 10 the amplification is 11%, which is where the threshold sits
The threshold is a calibrated judgement, not a physical boundary
An imperfection is replaced by H = φV, which produces exactly the same overturning
A second-order analysis on a perfect symmetric frame finds nothing to amplify

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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