CIV-01

Euler buckling load

Critical compressive load of a slender elastic column.

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StabilityEuler 1744EN 1993-1-1

Governing equation

P_{\mathrm{cr}} = \dfrac{\pi^2 E I}{(K L)^2}

where

E: Young modulus (GPa)
I: Second moment (cm⁴)
L: Length (m)
K: Effective-length factor (—)
P_{\mathrm{cr}}: Euler load (kN)

Lecture brief

Historical brief

From Euler’s 1744 elastica and Navier’s beam theory to Mohr’s circle and transformed-section RC, structural mechanics grew as a closed-form craft before finite elements. The lab keeps those governing lines for buckling, flexure, joints and influence. This sheet (CIV-01 — Euler buckling load) is the form associated with Euler 1744 · EN 1993-1-1. Working symbols: $E$ , $I$ , $L$ , $K$ $\rightarrow$ $P_{\mathrm{cr}}$ . Euler solved the elastica eigenvalue problem Pcr = π² EI /(KL)². The effective-length factor K encodes end restraint.

Purpose

Purpose: compute

P_{\mathrm{cr}}

from

E

I

L

K

in Structural & civil via

P_{\mathrm{cr}} = \dfrac{\pi^2 E I}{(K L)^2}

Critical compressive load of a slender elastic column. Use it when a real structural & civil question must be answered in SI before a code check.

Live realistic example

In symbols

Live case. Given

E = 200.000\,\mathrm{GPa}

I = 8000.000\,\mathrm{cm⁴}

L = 4.000\,\mathrm{m}

K = 1.000\,\mathrm{—}

, the governing relation

P_{\mathrm{cr}} = \dfrac{\pi^2 E I}{(K L)^2}

yields

P_{\mathrm{cr}} = 9869.604\,\mathrm{kN}

. Prismatic, initially straight, concentric load, remains elastic. Move a slider: the numbers are this situation, not a canned story.

Calculator

Inputs

Young modulus E200 GPaSecond moment I8000 cm⁴Length L4 mEffective-length factor K1 —

Outputs

Euler load P_{\mathrm{cr}}9869.604 kN

Reading speed

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Narration of this film

Prismatic, initially straight, concentric load, remains elastic.

Euler solved the elastica eigenvalue problem Pcr = π² EI /(KL)². The effective-length factor K encodes end restraint.

Reading speed

Watch on YouTube