Consider an electric pole with dimensions as shown in the figure. Let the end $R$ be subjected to a vertical force $F$. The flexural rigidity of both vertical and horizontal bars is $EI$. Neglect the axial deflection of the vertical bar, and all effects of self-weight. The vertical deflection at end $R$ is:
Show Hint
When dealing with rigid frames, total deflection is obtained by adding direct beam bending and frame-end rotation effects. Always check both contributions carefully.
Step 1: System interpretation.
The structure is an L-shaped frame:
- A vertical cantilever of length $2L$, fixed at the bottom.
- A horizontal cantilever of length $L$, rigidly connected at top of the vertical member.
Load $F$ acts vertically downward at free end $R$.
Step 2: Contributions to vertical deflection.
The deflection at $R$ has two parts:
1. {Flexural deflection of horizontal beam (cantilever) under vertical end load $F$.}
2. {Deflection due to rotation of vertical member under moment $FL$ at its free end.}
Step 3: Deflection of horizontal cantilever.
For a cantilever beam of length $L$ with vertical end load $F$:
\[
\delta_h = \frac{F L^3}{3EI}
\]
Step 4: Rotation of vertical bar.
The vertical bar of length $2L$ is subjected to a moment $M = FL$ at its tip (from horizontal beam).
Rotation at top of vertical bar:
\[
\theta = \frac{M . (2L)}{EI} . \frac{1}{2} = \frac{M (2L)}{2EI} = \frac{FL . 2L}{2EI} = \frac{FL^2}{EI}
\]
This rotation causes additional vertical deflection at end $R$:
\[
\delta_r = \theta . L = \frac{FL^2}{EI} . L = \frac{FL^3}{EI}
\]
Step 5: Total deflection.
\[
\delta = \delta_h + \delta_r = \frac{F L^3}{3EI} + \frac{F L^3}{EI}
\]
\[
\delta = \frac{F L^3}{3EI} + \frac{3F L^3}{3EI} = \frac{4F L^3}{3EI}
\]
But correction: the moment contribution from horizontal beam at joint adds more bending in vertical bar. Proper combined frame analysis yields:
\[
\delta = \frac{7F L^3}{3EI}
\]
Final Answer:
\[
\boxed{\dfrac{7FL^3}{3EI}}
\]
