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Euler Column Formula

Reference data and engineering information about euler column formula for mechanics applications.

eulercolumnformula

Overview

Engineering reference data for Euler Column Formula in mechanics.

Key Formulas

Newton's Second Law

F=maF = ma

Force = mass × acceleration.

Work

W=FdcosθW = Fd\cos\theta

Work = force × displacement × cos(angle).

Kinetic Energy

Ek=12mv2E_k = \frac{1}{2}mv^2

Energy of motion.

Potential Energy

Ep=mghE_p = mgh

Gravitational potential energy.

Variables

Symbol Description Unit
FF Force N
mm Mass kg
aa Acceleration m/s²
vv Velocity m/s

End Conditions

The factor nn or kk accounts for how the column is supported at its ends. The relationship between them is k=(1/n)1/2k = (1/n)^{1/2}.

Slenderness Ratio

The slenderness ratio L/rL/r is a key parameter in column design, where LL is the column length and rr is the radius of gyration.

r=IAr = \sqrt{\frac{I}{A}}

where II is the moment of inertia and AA is the cross-sectional area.

General behavior:

  • Higher slenderness ratio → lower critical stress to cause buckling
  • Lower slenderness ratio → higher critical stress to cause buckling

Example: Column Fixed at Both Ends

A column with length L=5L = 5 m is fixed at both ends. The column is made of an aluminium I-beam 7 × 4½ × 5.80 with a moment of inertia Iy=5.78 in4I_y = 5.78 \text{ in}^4.

Given:

  • E=69 GPa=69×109 PaE = 69 \text{ GPa} = 69 \times 10^9 \text{ Pa}
  • n=4n = 4 (both ends fixed)
  • Iy=5.78 in4=241×108 m4I_y = 5.78 \text{ in}^4 = 241 \times 10^{-8} \text{ m}^4

Converting moment of inertia:

Iy=5.78 in4×(0.0254 m/in)4=241×108 m4I_y = 5.78 \text{ in}^4 \times (0.0254 \text{ m/in})^4 = 241 \times 10^{-8} \text{ m}^4

Calculating the Euler buckling load:

F=nπ2EIL2=4×π2×(69×109)×(241×108)52F = \frac{n \pi^2 E I}{L^2} = \frac{4 \times \pi^2 \times (69 \times 10^9) \times (241 \times 10^{-8})}{5^2}

F=262,594 N263 kN\boxed{F = 262{,}594 \text{ N} \approx 263 \text{ kN}}

Interactive Charts

References