Leonhard Euler worked this out in 1744, and nothing since has improved on it for a slender column: $P_{cr} = \dfrac{\pi^{2} E I}{(KL)^{2}}$. Read aloud: P-crit equals pi squared E I over K L, all squared. Letter by letter — $P_{cr}$ is the critical buckling load in newtons, the axial load at which a perfectly straight column stops being straight; $E$ is Young's modulus, the material's stiffness (200 GPa for structural steel, 70 GPa for aluminium); $I$ is the least area moment of inertia of the section, because the column bows about its weakest axis; $K$ is the same effective length factor from the last lesson; and $L$ is the unbraced length. Whichever of those the question leaves blank is the one you solve for.

Read the shape of the relation before you read the numbers. The strength of the steel does not appear anywhere — only its stiffness. A slender column of high-strength steel buckles at exactly the same load as a slender column of mild steel, because E is practically identical for both. Paying for stronger steel in a slender column buys you nothing at all.

And the length sits in the denominator squared. Double the unbraced length and the capacity falls to a quarter; halve it with one mid-height brace and the capacity quadruples. That is why a brace is the cheapest structural steel you will ever specify. K, by the way, is a given on the paper and a judgement call in the field — it is never something to guess mid-problem.