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Euler Buckling Load of a Pinned Column

A slender steel column is pinned at both ends and has an unbraced length of $L = 3.0\ \text{m}$ .

Its material has $E = 200\ \text{GPa}$ and the section has a minimum second moment of area $I = 1.0 \times 10^{6}\ \text{mm}^4$ .

Find the Euler critical buckling load.

Given

L	3 m	Unbraced length
E	200 GPa	Young's modulus
I	1000000 mm^4	Minimum second moment of area

Hint 1

This is a stability problem, not a strength problem — the column fails by buckling, not by yielding.

Hint 2

P_cr = π²EI/(KL)². The whole question is what K should be.

Hint 3

Pinned at both ends means K = 1.0, so the effective length equals the actual length.

Worked solution — try the problem first

The Euler critical load is

$P_{cr} = \frac{\pi^2 E I}{(KL)^2}$

Effective length factor. For a column pinned at both ends, $K = 1.0$ , so $KL = L = 3.0\ \text{m}$ . (Fixed-fixed would give $K = 0.5$ ; fixed-free gives $K = 2.0$ .)

Convert units: $E = 200\times10^{9}\ \text{Pa}, \qquad I = 1.0\times10^{6}\ \text{mm}^4 = 1.0\times10^{-6}\ \text{m}^4$

Substitute: $P_{cr} = \frac{\pi^2 (200\times10^{9})(1.0\times10^{-6})}{(3.0)^2} = \frac{9.8696 \times 2.0\times10^{5}}{9}$

$= \frac{1.9739\times10^{6}}{9} = 2.193\times10^{5}\ \text{N}$

$\boxed{P_{cr} = 219\ \text{kN}}$

Note on the minimum I. Buckling always occurs about the axis of least stiffness, which is why the problem specifies the minimum second moment of area. Using the strong-axis value would badly overestimate capacity.

Concepts:Euler bucklingColumn stabilityEffective length factor

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