A cantilever of span L is subjected to a uniformly distributed load of intensity w from free end up to mid-span. Its deflection at the free end is
- A5wL⁴/384EI
- B41wL⁴/384EICorrect
- C21wL⁴/384EI
- D5wL⁴/192EI
Solution
Concept
Deflection by superposition / moment-area
Formula
δ = Σ (deflection of loaded part + slope × lever arm)
Step-by-step solution
1. UDL w acts over the free half (L/2). The unloaded fixed half carries V = wL/2 and M = wL²/8 at mid-span. 2. Deflection at mid-span = (wL/2)(L/2)³/3EI + (wL²/8)(L/2)²/2EI = wL⁴/48EI + wL⁴/64EI. 3. Slope at mid-span = (wL/2)(L/2)²/2EI + (wL²/8)(L/2)/EI = wL³/8EI; adds (wL³/8EI)(L/2) = wL⁴/16EI. 4. Free half itself: w(L/2)⁴/8EI = wL⁴/128EI. 5. Total = wL⁴ (8 + 6 + 24 + 3)/384EI = 41wL⁴/384EI.
Explanation
The tip deflection is the sum of the mid-span deflection, the effect of the mid-span slope over the free half, and the deflection of the loaded half as a small cantilever.
Why the other options are wrong
5wL⁴/384EI is the simply supported beam value; 21wL⁴/384EI and 5wL⁴/192EI do not equal the sum 41/384.
Shortcut
Break the cantilever at mid-span and add three parts.
Common mistake
Using w L⁴/8EI for the full length.
Correct answer: B) 41wL⁴/384EI