Module 18 · Lesson 18.5

Beam-columns: bending and buckling together

What happens when a member carries transverse load and axial compression at once.

Why this matters

Module 9 added bending and axial stresses by superposition, and flagged that it fails for slender members. This is the repair. A member carrying both actions bends more than the transverse load alone would cause, because the axial force acts through the deflection it has already produced — and that feedback is what connects the bending problem to the buckling one.

By the end of this lesson you should be able to

Explain the second-order (P-delta) effect in physical terms
Use the moment magnification factor
Recognise when superposition is adequate and when it is not
Describe the secant formula and what it adds

What you should already know

Euler's critical load and effective length (this module)
Combined bending and axial load (Module 9)
Beam deflection (Module 13)

Take a member with a transverse load. It deflects by some amount δ₀ and carries a first-order moment M₀. Now add an axial compression P.

That axial force is now acting at an offset δ₀ from the line between the supports, so it applies an extra moment Pδ₀. Extra moment means extra deflection. Extra deflection means a longer lever arm for P, so more moment again — and round it goes.

For an elastic member the series converges, and the sum is remarkably tidy:

M_max = M₀ / (1 − P/P_cr)

The quantity 1/(1 − P/P_cr) is the moment magnification factor, and it is one of the most useful numbers in structural engineering.

Moment magnification

M_{ma x} = \frac{M _{0}}{1 - P / P _{cr}}

What it calculates: the second-order moment in a member carrying axial compression

M₀: first-order moment, from the transverse load alone (kNm)
P: applied axial compression (kN)
P_cr: Euler critical load for the member (kN)

This assumes

Elastic behaviour throughout
Compression — a tensile axial force reduces the moment instead
The deflected shape is close to the buckling mode, which it usually is

In plain terms: At P = 0 the factor is 1 and superposition is exact. At P = P_cr it is infinite — which is the physical meaning of the critical load for a member that is already bent: no sudden bifurcation, just deflections growing without limit.

The factor gives a clean way to judge when superposition is good enough:

P/P_cr	Magnification	Verdict
0.1	1.11	11% — often acceptable
0.25	1.33	33% — must be included
0.50	2.00	The moment doubles
0.75	4.00	Four times
0.90	10.0	The member is effectively gone

Notice how sharply it climbs. The relationship is not gentle, and a member at 0.75 of its critical load has almost no useful bending capacity left.

This also explains the reverse effect: an axial tension produces a magnification factor of 1/(1 + P/P_cr), which is less than one. Tension straightens a member and reduces its moment. It is why a tie rod under load is stiffer than the same rod slack, and why prestressing a member in tension is a legitimate way to stiffen it.

Predict first

A beam-column carries an axial load equal to half its Euler critical load. By how much is its first-order moment magnified?

One refinement remains. The magnification factor assumes the transverse loading produces a deflected shape close to the buckling mode. For an initially crooked or eccentrically loaded strut, the secant formula gives an exact elastic answer:

σ_max = (P/A)[1 + (ec/r²)·sec( (L_e/2r)·√(P/EA) )]

The term ec/r² is the eccentricity ratio, and the secant carries the amplification. Two things are worth noting:

With e = 0 it reduces to σ = P/A, as it must.
As the argument of the secant approaches π/2, the secant goes to infinity — and that happens precisely when P reaches the Euler load.

The secant formula is exact but awkward to invert: you cannot rearrange it for P. That is why it is used to understand behaviour and to generate design curves, rather than as a working design formula.

Worked example

A beam-column carrying wind and gravity

Given

Pin-ended column, 4.00 m long, E = 205 000 N/mm², I = 4.00 × 10⁶ mm⁴, A = 3000 mm²
Transverse load produces a first-order mid-height moment M₀ = 25.0 kNm
Axial compression P = 250 kN

Find

The Euler load, the magnification factor and the second-order moment.

Practice

A pin-ended column 4.00 m long has E = 205 000 N/mm² and I = 4.00 × 10⁶ mm⁴. What is its Euler critical load, in kN?

Practice

That column carries 250 kN of axial compression. What is the moment magnification factor?

Practice

If the first-order moment is 25.0 kNm, what is the second-order moment, in kNm?

Summary

Axial compression acting through an existing deflection adds moment, which adds deflection, and so on
The elastic series sums to M_max = M₀/(1 − P/P_cr)
At P/P_cr = 0.5 the moment doubles; at 0.9 it is magnified tenfold
The error from ignoring it is always unsafe
Axial tension gives 1/(1 + P/P_cr), which straightens the member and reduces its moment
The secant formula is the exact elastic result for an eccentrically loaded strut, but cannot be inverted for P

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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