# Euler's Column Formula

## Buckling of columns

Columns fail by buckling when their critical load is reached. Long columns can be analysed with the Euler column formula

*F = n π ^{2} E I / L^{2} (1) *

*where *

*F = allowable load (lb, N)*

*n = factor accounting for the end conditions*

*E = modulus of elastisity (lb/in ^{2}, Pa (N/m^{2}))*

*L = length of column (in, m)*

*I = Moment of inertia (in ^{4}, m^{4})*

### Factor Counting for End Conditions

- column pivoted in both ends :
*n = 1* *both ends fixed :**n = 4*- one end fixed, the other end rounded :
*n = 2* - one end fixed, one end free :
*n = 0.25*

### Note!

Equation *(1)* is sometimes expressed with a k factor accounting for the end conditions:

*F = π ^{2} E I / (k L)^{2} (1b)*

*where*

*k = (1 / n) ^{1/2} factor accounting for the end conditions*

n | 1 | 4 | 2 | 0.25 |

k | 1 | 0.5 | 0.7 | 2 |

### Example - A Column Fixed in both Ends

An column with length *5 m* is fixed in both ends. The column is made of an Aluminium I-beam 7 x 4 1/2 x 5.80 with a Moment of Inertia *i _{y} = 5.78 in^{4}*. The Modulus of Elasticity of aluminum is

*69 GPa (69 10*and the factor for a column fixed in both ends is

^{9}Pa)*4.*

The Moment of Inertia can be converted to metric units like

*I _{y} = 5.78 in^{4} (0.0254 m/in)^{4 }*

* = 241 10 ^{-8} m^{4}*

The Euler buckling load can then be calculated as

*F = 4 π ^{2} (241 10^{-8} m^{4}) (69 10^{9} Pa) / (5 m)^{2} *

* = 262594 N*

* = 263 kN*

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