What is the horizontal sway of the point 'B' of the portal frame shown in the figure given below :

\frac{P L^{3}}{15 E I}

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\frac{P L^{3}}{18 E I}

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\frac{P L^{3}}{24 E I}

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\frac{P L^{3}}{12 E I}

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Solution

The correct option is B $\frac{P L^{3}}{18 E I}$

$\frac{12 E I δ}{L^{3}} + \frac{6 E I δ}{L^{3}} = P$

$\Rightarrow δ = \frac{P L^{3}}{18 E I}$

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