CIV-13

Euler buckling load

Pcr = π² EI / (K L)². The first eigenvalue of a slender column.

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StabilityEuler

Governing equation

P_{cr}=\dfrac{\pi^2 EI}{(KL)^2}

where

E: Modulus (GPa)
I: Second moment (cm^4)
L: Length (m)
K: Effective length factor (—)
P_{cr}: Critical 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-13 — Euler buckling load) is the form associated with Euler. Working symbols: $E$ , $I$ , $L$ , $K$ $\rightarrow$ $P_{cr}$ . K encodes the end restraints: 1 pinned–pinned, 0.5 fixed–fixed, 2 cantilever.

Purpose

Purpose: compute

P_{cr}

from

E

I

L

K

in Structural & civil via

P_{cr}=\dfrac{\pi^2 EI}{(KL)^2}

Pcr = π² EI / (K L)². The first eigenvalue of a slender 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 = 800.000\,\mathrm{cm^4}

L = 4.000\,\mathrm{m}

K = 1.000\,\mathrm{—}

, the governing relation

P_{cr}=\dfrac{\pi^2 EI}{(KL)^2}

yields

P_{cr} = 987.0\,\mathrm{kN}

. A slender strut, a sine half-wave, a critical load. Move a slider: the numbers are this situation, not a canned story.

Calculator

Inputs

Modulus E200 GPaSecond moment I800 cm^4Length L4 mEffective length factor K1 —

Outputs

Critical load P_{cr}987.0 kN

Reading speed

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Narration of this film

A slender strut, a sine half-wave, a critical load.

K encodes the end restraints: 1 pinned–pinned, 0.5 fixed–fixed, 2 cantilever.

Reading speed

Watch on YouTube