A single bay portal frame of height h fixed at the base is subjected to a horizontal displacement of δ at the top. Assuming all members to be prismatic, the base moments developed is proportional to:

This question was previously asked in

CIL MT Civil : 2020 Official Paper

Option 1 : \(\frac{1}{{{h^2}}}\)

For the given frame:

Let, the horizontal displacement of column CD = δ.

Taking moments about ‘C’ & using moment area theorem,

∑ A_{i} x_{i}/EJ = -δ

\({{\rm{A}}_1} = - \left( {\frac{1}{2}} \right)\left( {\frac{{\rm{l}}}{2}} \right)\left( {\rm{M}} \right) = \frac{{ - {\rm{Ml}}}}{4}\)

\({{\rm{A}}_2} = \left( {\frac{1}{2}} \right)\left( {\frac{{\rm{l}}}{2}} \right)\left( {\rm{M}} \right) = \frac{{{\rm{Ml}}}}{4}\)

\({{\rm{x}}_1} = \frac{1}{2} + \frac{2}{3}\left( {\frac{{\rm{l}}}{2}} \right) = \frac{{5{\rm{l}}}}{6}\)

\({{\rm{x}}_2} = \frac{1}{3} \times \frac{{\rm{l}}}{2} = \frac{{\rm{l}}}{6}\)

\(\frac{1}{{{\rm{Ej}}}}\left[ { - \left( {\frac{{{\rm{Ml}}}}{4}} \right)\left( {\frac{{5{\rm{l}}}}{6}} \right) + \left( {\frac{{{\rm{Ml}}}}{4}} \right)\left( {\frac{{\rm{l}}}{6}} \right)} \right] = - {\rm{\sigma }}\)

\(\left( {\frac{{ - 5{\rm{M}}{{\rm{l}}^2}}}{{24}} + \frac{{{\rm{M}}{{\rm{l}}^2}}}{{24}}} \right)\frac{1}{{{\rm{EJ}}}} = - {\rm{\sigma }}\)

\({\rm{M}} = \frac{{6{\rm{EJ\sigma }}}}{{{{\rm{l}}^2}}}\)

\({\rm{M}} \propto \frac{1}{{{{\rm{l}}^2}}}\)

∵ ℓ = h

\(\therefore {\rm{M}} \propto \frac{1}{{{{\rm{h}}^2}}}\)