Question

For long shallow girders with low warping stiffness the critical moment is expressed as
1. \(\sqrt {{\rm{E}}{{\rm{I}}_{\rm{y}}}{\rm{GJ}}} \)
2. \(\frac{{\rm{\pi }}}{{\rm{L}}}\sqrt {{\rm{E}}{{\rm{I}}_{\rm{y}}}{\rm{GJ}}} \)
3. \(\frac{{\rm{\pi }}}{{\rm{L}}}\left( {{\rm{E}}{{\rm{I}}_{\rm{y}}}{\rm{GJ}}} \right)\)
4. \({\rm{E}}{{\rm{I}}_{\rm{y}}}{\rm{GJ}}\)

Answer 1

Correct Answer - Option 2 : \(\frac{{\rm{\pi }}}{{\rm{L}}}\sqrt {{\rm{E}}{{\rm{I}}_{\rm{y}}}{\rm{GJ}}} \)

Critical moment:

If symmetrical I-beam is subjected to couple M, at the ends and beam is restrained against torsion at the ends, the lateral buckling of beam takes place when the value of M attains critical value.

The critical moment M is given by,

\({{\rm{M}}_{{\rm{cr}}}} = {\left[ {{\rm{E}}{{\rm{I}}_{\rm{y}}}{\rm{C}} \times \left( {1 + \left( {\frac{{{{\rm{C}}_1}{{\rm{\pi }}^2}}}{{{{\rm{C}}_2}{{\rm{l}}^2}}}} \right)} \right)} \right]^{\frac{1}{2}}}\)

For long shallow girders with low warping stiffness the critical moment is expressed as

\({{\rm{M}}_{{\rm{cr}}}} = \frac{{\rm{\pi }}}{{\rm{L}}}\sqrt {{\rm{E}}{{\rm{I}}_{\rm{y}}}{\rm{GJ}}} \)

Where,

L = Effective length of compression flange,

C = Torsional rigidity of beam,

EI_y = flexural rigidity about week axis, and

GJ = Torsional rigidity