In beam-columns or eccentric loaded columns, an elastic critical stress in compression f_{cc} is

Where: E = Modulus of elasticity of steel

λ = Slenderness ratio in the plane of bending

This question was previously asked in

ESE Civil 2020: Official Paper

Option 2 : \(\frac{{{\pi ^2}E}}{{{\lambda ^2}}}\)

CT 1: Current Affairs (Government Policies and Schemes)

55064

10 Questions
10 Marks
10 Mins

__Concept: __

Elastic critical load in columns is given by Euler’s Formula which is, \({{\rm{P}}_{{\rm{cr}}}} = \frac{{{{\rm{\pi }}^2}{\rm{EI}}}}{{{\rm{L}}_{\rm{e}}^2}}\)

Where, \({{\rm{P}}_{{\rm{cr\;}}}}\)is the critical load of the column, E is the modulus of elasticity of the column, I is the moment of inertia of the column cross section, \({{\rm{L}}_{{\rm{e\;}}}}\) is the effective length of the column.

Now, effective length (Le) for different end conditions in terms of actual length (L) are listed in the following table:

Support Conditions |
Effective length (Le) |

Both ends hinged/pinned |
Le = L |

One end hinged other end fixed |
Le = L/√2 |

Both ends fixed |
Le = L/2 |

One end fixed and other end free |
Le = 2L |

Now, moment of inertia can be expressed as, \(I = {\rm{A}}{{\rm{r}}^2}\) where A is the cross sectional area of the column and r is the radius of gyration for the column cross section.

\(\therefore {\rm{\;}}{{\rm{P}}_{{\rm{cr}}}} = \frac{{{{\rm{\pi }}^2}{\rm{EI}}}}{{{\rm{L}}_{\rm{e}}^2}} = \frac{{{{\rm{\pi }}^2}{\rm{EA}}{{\rm{r}}^2}}}{{{\rm{L}}_{\rm{e}}^2}} = \frac{{{{\rm{\pi }}^2}{\rm{EA}}}}{{{{\left( {\frac{{{{\rm{L}}_{\rm{e}}}}}{{\rm{r}}}} \right)}^2}}} = \frac{{{{\rm{\pi }}^2}{\rm{EA}}}}{{{{\left( {\rm{\lambda }} \right)}^2}}}{\rm{\;\;}}\) where λ is slenderness ratio.

∴ Elastic critical stress, \({\rm{\;}}{{\rm{f}}_{{\rm{cc}}}} = \frac{{{{\rm{P}}_{{\rm{cr}}}}}}{{\rm{A}}} = \frac{{{{\rm{\pi }}^2}{\rm{E}}}}{{{{\left( {\rm{\lambda }} \right)}^2}}}\)