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Structural Analysis Fundamentals / Structural instability

Beam-columns: bending and buckling together

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

Lesson review: expert verification remains open

This lesson has received a full-context editorial review and a check of its technical explanations. Selected arithmetic has been checked separately. This is not independent engineering sign-off.

Checks needed for project use

  • An independent structural reviewer must check remaining derivations, diagrams, sign conventions and the interactive models across their full input ranges.
  • Independent stability review must cover imperfections, local/flexural-torsional modes, restraint, material nonlinearity and code resistance; ideal elastic examples are not design capacities.

These are educational models and exercises. Check their assumptions before using the results. The full technical content and every diagram have not yet been independently verified.

Concrete and steel calculations retain their stated first-generation Eurocode teaching basis. Use the standards and amendments specified for your project; do not mix generations without an appropriate design basis.

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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:

Mmax = M₀ / (1 − P/Pcr)

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

Moment magnification
Mmax=M01−P/PcrM_{max} = \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)
Pcr
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 = Pcr 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 first-mode estimate gives magnification factors of 1.11, 1.33, 2.00, 4.00 and 10.0 for P/Pcr of 0.10, 0.25, 0.50, 0.75 and 0.90. Whether a change is acceptable depends on the required accuracy and design checks. These ratios do not by themselves establish remaining bending resistance.

The estimate assumes an appropriate mode shape and 0 ≤ P < Pcr. Different load distributions and restraints can require a full second-order analysis. Axial tension tends to reduce transverse displacement in the corresponding model.

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?

For an initially straight, prismatic pin-ended column with equal end eccentricity e, linear elasticity and small rotations, the secant solution is

σmax = (P/A)[1 + (ec/r²) sec((L/2r)√(P/EA))]

It applies below the ideal Euler load. Here L is the actual pin-to-pin length. Arbitrary crookedness or end restraint cannot be represented by substituting an effective length without establishing equivalence. A sinusoidal initial bow has a different amplification expression.

At e = 0, the straight pre-buckling branch gives P/A; the separate stability check remains necessary.

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 Mmax = M₀/(1 − P/Pcr)
    • At P/Pcr = 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/Pcr), which straightens the member and reduces its moment
    • The secant formula describes the stated initially straight pin-ended eccentric column; solve numerically for a chosen stress limit and check applicability