Euler Column Buckling

P_{cr} = \frac{π ^{2} E I}{( K L ) ^{2}}

Elastic critical load for a long straight column. K is the effective-length factor (pinned-pinned K=1, fixed-fixed 0.5, fixed-free 2.0, fixed-pinned 0.7 theoretical). Use I about the weak axis unless braced. Inelastic columns use tangent modulus / AISC curves, not this raw Euler load.

$P_{cr}$ (N): Critical load
$K$: Effective length factor
$L$ (m): Unbraced length
$E, I$: Modulus and inertia

Step-by-step solved example

Pinned-pinned steel, L = 4 m, I = 4.0×10⁷ mm⁴, E = 200 GPa. Pcr?

Same Euler formula as mechanics; structural K from recommended values.

1. SI
I = 4.0×10⁻⁵ m⁴, KL = 4 m.
2. Euler
Pcr = π² (200e9)(4.0e-5) / 16 = 4.93×10⁶ N ≈ 4930 kN.
$P_{cr} \approx 4930 kN$

Answer: about 4930 kN

10 practice questions

0/10 correct

1.Pinned-pinned K =

2.Fixed-free (cantilever column) K =

3.Doubling L multiplies Pcr by

4.Use the smaller I because

5.Slenderness KL/r large means

6.Recommended design K for fixed-fixed is often

7.Bracing the midpoint about the weak axis

8.Euler stress Fcr = Pcr/A =

9.If Fcr > Fy Euler is

10.Units: E in kPa, I in m⁴, L in m → Pcr in