A cantilever of span l is fixed at one end, and its other end rests freely on the middle of a simply supported cross-beam of the same span and section. If the cantilever carries a UDL of w per unit length, the reaction at its free end offered by the cross-beam is:
Try it first: pick an option, then read the explanation.
Asked in SSC JE 23/03/2021 (Morning)
Detailed Explanation
The two beams touch at one point, so they must deflect by the same amount there (compatibility). Let R be the contact force: it pushes the cantilever tip up and the cross-beam centre down.
Cantilever tip: \( \delta_1 = \dfrac{wl^4}{8EI} - \dfrac{Rl^3}{3EI} \) (downward). Cross-beam centre: \( \delta_2 = \dfrac{Rl^3}{48EI} \).
\[ \frac{wl^4}{8EI} - \frac{Rl^3}{3EI} = \frac{Rl^3}{48EI} \Rightarrow \frac{wl}{8} = R\left(\frac{16+1}{48}\right) \Rightarrow R = \frac{48wl}{136} = \frac{6wl}{17} \]
Why “6wl/17” Is Correct
Solving the compatibility equation gives R = 6wl/17 ≈ 0.353wl. If the support were rigid, R would be the propped-cantilever value 3wl/8 = 0.375wl; the flexible cross-beam yields a little, so R is slightly smaller.
Why the Other Options Are Wrong
- A. 3wl/17: 3wl/17 is half the correct value.
- C. 3wl/8: 3wl/8 is the prop reaction for a rigid (unyielding) prop; here the prop is a flexible beam.
- D. 2wl/5: 2wl/5 has no basis in the compatibility equation.
Key Point
Two beams in contact: equal deflection at the contact point. Here wl⁴/8 − Rl³/3 = Rl³/48 ⇒ R = 6wl/17.
Key Concepts for Students
- Rigid prop under a UDL cantilever: R = 3wl/8.
- Flexible supports reduce the prop reaction.
- Superposition + compatibility is the basis of the force method.
Practice more: this is question 14 in Slope and Deflection MCQs – Part 1. Try the whole set and check every answer instantly.