Module 9 · Lesson 9.1

Why N and M do not simply add

The axial force does not just eat into the moment capacity. It also increases the moment — and that second effect is the one people miss.

Why this matters

A column with a beam framing into one side carries axial force and moment together. The instinct is to check each, add the utilisations, and stop when the sum reaches one. That instinct is right about the first effect and completely blind to the second: the moment bows the member, and the axial force then acts through that bow and produces more moment. On a stocky column the extra is a few percent and the instinct survives. On a slender one it is the difference between a member that passes and a member that fails.

By the end of this lesson you should be able to

Explain the feedback loop between moment and axial force in a member
Derive the amplification factor and state what it depends on
Say why grade does not help and stiffness does
Separate the three things all called 'the interaction check'

What you should already know

The Euler load and relative slenderness (Module 7)
Cross-section N–M interaction, m = 1 − n² for a rectangle (Module 6)
Lateral-torsional buckling and M_b,Rd (Module 8)
The magnification series for P–Δ effects (Module 4)

Two separate effects, often confused

Put an axial force and a moment on the same member and two quite different things happen.

The first is a cross-section effect. Some of the section's material is used carrying the axial force, so less is available for bending. For a rectangle this is exactly m = 1 − n², which Module 6 derived from where the plastic neutral axis sits. It is a plasticity result, it needs no calibration, and it is entirely local — it would apply to a 10 mm long stub.

The second is a member effect, and it is not local at all. The moment bends the member, so its centreline is no longer straight. The axial force now acts along a line that is offset from the deflected centreline, and an offset force produces a moment. That extra moment bends the member further, which increases the offset, which increases the moment again.

That loop either settles down or it does not. Where it settles is what the amplification factor tells you.

The first effect is about the section. The second is about the member. A check that only does the first is checking the wrong thing.

From first principles

The amplification factor

We want to show: Show that a first-order moment M₀ becomes M₀/(1 − N/N_cr) once the member's own deflection is accounted for, and see what that expression does and does not contain.

Bend the member and it deflects. Push on the ends and that force now acts through the deflection, adding moment, which adds deflection, which adds moment. Each round is smaller than the last — provided the axial force is below the critical load. Add up the rounds and you get a finite answer. The closer N gets to N_cr, the more slowly the rounds shrink, and the larger the total.

Take care

Three different things are all called 'the interaction check'

Keeping these apart is most of what makes beam-column design comprehensible.

1. Cross-section interaction — m = 1 − n² for a rectangle, or the calibrated expression for an I section. A plasticity result about a section. Module 6.

2. Member amplification — M = M₀/(1 − N/N_cr). Derived above. Mechanics, about a member.

3. The Eurocode member buckling interaction — the two expressions with their k factors. These are code-calibrated. They fold amplification, moment shape, lateral-torsional buckling and cross-section plasticity into a form fitted against numerical and test results, and the Eurocode offers two alternative derivations of the factors whose selection is a national choice.

This course teaches (1) and (2) in full, because they are derivable and they are what the third is calibrated to reproduce. It does not reproduce (3), and the simplified expression used here is labelled as a teaching simplification and is disabled for design use. For a real check, go to the published standard and its National Annex.

Try it

The straight line, and what it misses

The vertical axis is axial utilisation, the horizontal axis is how much bending capacity is left. The straight line is the simple check. The curve is what happens once the member's own deflection is counted. Change the length and watch the gap open.

Member length12.0 m

Axial utilisation0.500

Moment shape

Member length: 12.0 m
Critical load N_cr: 2029 kN
Buckling resistance N_b,Rd: 1518 kN
Slenderness λ̄_y: 1.399
Axial force N_Ed: 759 kN
Ratio N_Ed/N_cr: 0.374
Amplification: 1.597
Moment factor C_m: 1.00
Straight line allows: 50 %
Amplified check allows: 31 %
Lost to amplification: 37 %

The straight line allows 50% of the bending capacity here; the amplified check allows 31%. That is 37% of the moment capacity the simple check would have handed you and the member does not have. N_Ed/N_cr = 0.374 is what drives it.

Things worth trying

Start at 6 m with uniform moment and the axial utilisation at 0.5. The readout says 19% — that share of the bending capacity the straight line offers is not actually there. Real, but survivable.
Now switch to single curvature at the same point: 10%. C_m = 0.9 has taken about half of it back, which is exactly what happens in this module's first worked example, where the two checks came out at 0.601 and 0.608.
Stay on single curvature and take the axial utilisation down to 0.1. The figure goes NEGATIVE and the curve crosses above the line — with a favourable moment shape and a light axial load, the straight line is conservative. It is not reliably safe or reliably unsafe, which is the real objection to it.
Back to uniform moment, and take the length to 12 m at n = 0.5: 37%. Nothing about the section changed; only N_cr, which goes with 1/L².
The straight line still offers half the bending capacity there. The amplified curve allows about a third.
Push the length to 16 m and the utilisation to 0.8. Two thirds of the straight line's allowance has gone, and no cross-section calculation would have told you.
Switch to double curvature at 6 m. The curve now sits ABOVE the straight line at low axial load, because C_m = 0.4 more than cancels a small amplification. The simple check is conservative there — which is worth knowing, because it means the straight line is not reliably safe OR reliably unsafe.
Watch λ̄_y and the amplification together as you change the length. They rise together, and that is the compounding that makes slender beam-columns dangerous: the axial resistance is falling at the same time as the moment demand is rising.

Worked example

A 6 m braced column, restrained at mid-height

Given

254 × 254 × 89 UC in S355: A = 111.9 cm², I_y = 14100 cm⁴, I_z = 4857 cm⁴, W_pl,y = 1209 cm³
Storey height 6.0 m, pinned top and bottom for major-axis bending
A tie at mid-height restrains the minor axis, so L_cr,z = 3.0 m
N_Ed = 1000 kN with a first-order moment M_Ed = 120 kNm about the major axis, single curvature
Class 1 in both compression and bending

Find

Whether the column is adequate, and how much of the final moment the column created itself.

Assumptions

Buckling curves b (major) and c (minor) for a rolled H section — calibrated and unverified here
C_m = 0.9 for end moments in single curvature — calibrated
The interaction used is the course's teaching simplification, not the code expression

Worked example

The same section, 12 m tall — where the straight line fails

Given

The same 254 × 254 × 89 UC in S355
12.0 m between restraints about the major axis, pinned both ends
N_Ed = 900 kN with a first-order moment M_Ed = 120 kNm, single curvature
Assume the minor axis and LTB are restrained and do not govern, so the in-plane check is isolated

Find

Whether the straight-line check and the amplified check agree.

Assumptions

Same calibrated buckling curve and C_m as before
The minor axis is assumed restrained purely to isolate the effect being demonstrated — a real 12 m column would need that restraint checked

Predict first

The 12 m column that failed is respecified in a higher grade, S460 rather than S355, keeping the same section. What happens to the amplified check?

Practice

A member carries N_Ed = 900 kN and has N_cr = 2029 kN in the plane of bending. What is the amplification factor on its first-order moment?

Practice

A first-order moment of 120 kNm is amplified by 1.797 and then corrected by C_m = 0.9. What is the design moment, in kNm?

Practice

A column carries N_Ed = 1000 kN and has N_cr,y = 8118 kN. What is N_Ed/N_cr,y?

Practice

At what value of N/N_cr does the amplification factor reach exactly 2?

Check yourself

Why does a higher steel grade do almost nothing for the amplification of a beam-column's moment?

Summary

N and M interact twice: once in the section, and once through the member's deflection
M = M₀/(1 − N/N_cr), derived from equilibrium on the deflected shape
At N/N_cr = 0.5 the moment doubles — the relationship is strongly non-linear
The amplification contains E, I and L but not f_y, so grade does not help it
N/N_cr is NOT N/N_b,Rd — in the first worked example they were 0.12 and 0.32
C_m corrects for moment shape and is calibrated, like C₁ for beams
A 6 m column: straight line 0.601, amplified 0.608 — the corrections nearly cancelled
The same section at 12 m: straight line 0.873, amplified 1.045 — it fails

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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