Euler Buckling Load

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

Critical concentric load for a long, slender, linearly elastic column. K is the effective-length factor: 1.0 pinned–pinned, 0.5 fixed–fixed, 0.7 fixed–pinned, 2.0 fixed–free. Use the minimum centroidal I. Short columns yield before Euler applies.

$P_{cr}$ (N): Euler critical load
$E$ (Pa): Modulus of elasticity
$I$ (m⁴): Minimum centroidal moment of inertia
$K$ (—): Effective-length factor
$L$ (m): Unbraced length

Step-by-step solved example

Pinned–pinned (K = 1) steel column, L = 2.0 m, I = 4.0×10⁻⁸ m⁴, E = 200 GPa. Pcr?

Pcr = π²EI/(KL)². K from end conditions (pinned, fixed, free).

1. EI and π²
EI = 200e9 × 4.0e−8 = 8000 N·m². π² EI = 7.90×10⁴.
2. Divide by (KL)²
(KL)² = 4.00. Pcr = 7.90×10⁴ / 4 = 19.7 kN.
$P_{cr} = 19.7 kN$

Answer: 19.7 kN

10 practice questions

0/10 correct

1.Pinned–pinned K is

2.Pcr is proportional to

3.Fixed–free (cantilever) recommended K is

4.Fixed–fixed recommended K is

5.Doubling L (same K, E, I) divides Pcr by

6.Euler’s formula assumes

7.I in Pcr should be

8.Increasing E

9.K = 1, L = 2 m, EI = 8.00×10⁴ N·m². Pcr ≈

10.Short stocky columns typically fail by