The deflection of the free end of a cantilever of span L having a concentrated load W at mid-span is:
Try it first: pick an option, then read the explanation.
Correct answer: C. 5WL³/(48EI)
Asked in SSC JE 2013
Detailed Explanation
The half beyond the load carries no moment and stays straight, so the tip deflection = deflection under the load + slope at the load × L/2.
\[ y_B = \frac{W(L/2)^3}{3EI} = \frac{WL^3}{24EI},\qquad \theta_B = \frac{W(L/2)^2}{2EI} = \frac{WL^2}{8EI} \]
\[ y_C = \frac{WL^3}{24EI} + \frac{WL^2}{8EI}\cdot\frac{L}{2} = \frac{2WL^3 + 3WL^3}{48EI} = \frac{5WL^3}{48EI} \]
Why “5WL³/(48EI)” Is Correct
The free-end deflection is 5WL³/(48EI).
Why the Other Options Are Wrong
- A. WL³/(3EI): WL³/3EI applies only when W is at the free end.
- B. 5WL³/(24EI): 5WL³/24EI is twice the correct value.
- D. WL³/(48EI): WL³/48EI is the simply supported central-load result.
Key Point
Cantilever, W at mid-span: δ under load = WL³/24EI, δ at tip = 5WL³/48EI.
Key Concepts for Students
- General: tip deflection = Wa²(3L − a)/6EI.
- Slope at the tip equals the slope under the load (WL²/8EI).
- Straight unloaded segments simplify cantilever problems.
Practice more: this is question 36 in Slope and Deflection MCQs – Part 3. Try the whole set and check every answer instantly.