A column: which buckling formula, and why the answer changes at a slenderness

A short column crushes and a long one buckles, and the two have entirely different formulas. Which applies is not a judgement call — it is decided by the slenderness ratio against a transition value that follows from the material, and getting it wrong overestimates a long column's capacity by a factor that grows without limit. This is a solid round strut: 50 mm diameter, 2 m long, pinned at both ends, S355, carrying 40 kN in compression.

use steel

Four significant figures.

digits 4

d

50 mm

L

2 m

P

40 kN

The effective length factor. Pinned at both ends is the reference case and the one that is safe to assume when the real restraint is uncertain: a fixed end that works loose becomes a pin, and a pin never becomes a fixed end.

K_{eff}

1

The section

A

\frac{π \cdot d^{2}}{4}

\frac{π \cdot {(50 mm)}^{2}}{4}

1963 {mm}^{2}

I

\frac{π \cdot d^{4}}{64}

\frac{π \cdot {(50 mm)}^{4}}{64}

306800 {mm}^{4}

r_{gyr}

\sqrt{\frac{I}{A}}

\sqrt{\frac{306800 {mm}^{4}}{1963 {mm}^{2}}}

12.5 mm

Slenderness, and the transition

The transition is where Euler's curve meets the parabola through the yield point — below it the column is in the inelastic range and Euler overstates what it can carry, above it Euler is the answer.

λ

\frac{K_{eff} \cdot L}{r_{gyr}}

\frac{1 \cdot 2 m}{12.5 mm}

160

λ_{c}

\sqrt{\frac{2 \cdot π^{2} \cdot E_{steel}}{{Fy}_{S355}}}

\sqrt{\frac{2 \cdot π^{2} \cdot 200 GPa}{355 MPa}}

105.5

{is}_{long}

λ \geq λ_{c}

160 \geq 105.5

1

The critical stress

One expression with the decision inside it, so the worksheet cannot be read as having chosen a formula and forgotten to say which.

σ_{cr}

{\begin{cases} \frac{π^{2} \cdot E_{steel}}{λ^{2}} & if {is}_{long} \\ {Fy}_{S355} - \frac{{\frac{{Fy}_{S355} \cdot λ}{2 \cdot π}}^{2}}{E_{steel}} & otherwise \end{cases}

{\begin{cases} \frac{π^{2} \cdot 200 GPa}{160^{2}} & if 1 \\ {Fy}_{S355} - \frac{{\frac{{Fy}_{S355} \cdot λ}{2 \cdot π}}^{2}}{E_{steel}} & otherwise \end{cases}

77110000 Pa

σ_{cr}

77.11 MPa

P_{cr}

σ_{cr} \cdot A

77110000 Pa \cdot 1963 {mm}^{2}

151.4 kN

The check

Three against buckling rather than the two a yield check takes. Buckling is sudden, it has no ductile reserve past the peak, and the load that causes it depends on straightness and end restraint that nobody measures.

n_{buckling}

3

check

P \leq \frac{P_{cr}}{n_{buckling}}

40 kN \leq \frac{151.4 kN}{3}

pass

Critical stress against slenderness

Both regimes on one curve, with the corner at the transition. The flat part on the left is the parabola approaching yield; the falling part on the right is Euler.

fn critical defined

plot (critical, 20, 250)

What this leaves out

A perfectly straight, centrally loaded, prismatic strut. A real one is none of those, and the codes handle it with a column curve that folds initial bow, residual stress and eccentricity into a reduction factor rather than with the Euler load and a factor of safety. This is the calculation that shows what is happening; a code check is what a drawing is signed against.