For a cantilever of length L carrying a triangular load of intensity W at the support and zero at the free end, the slope at the free end is:
Try it first: pick an option, then read the explanation.
Correct answer: C. WL³/(24EI)
Asked in SSC JE 2012
Detailed Explanation
With M = −W(L − x)³/(6L), one integration with zero slope at the fixed end gives the free-end slope.
\[ EI\theta_B = \int_0^L \frac{W(L-x)^3}{6L}dx = \frac{W}{6L}\cdot\frac{L^4}{4} = \frac{WL^3}{24} \]
Why “WL³/(24EI)” Is Correct
The free-end slope is WL³/(24EI).
Why the Other Options Are Wrong
- A. WL³/(8EI): WL³/8EI is the slope when the triangle's peak is at the free end.
- B. WL³/(12EI): WL³/12EI is not a standard result.
- D. WL³/(48EI): WL³/48EI is too small.
Key Point
Triangular load, peak at fixed end: θ = WL³/24EI, δ = WL⁴/30EI.
Key Concepts for Students
- Mohr I: area of M/EI diagram = WL³/24EI here.
- UDL cantilever: θ = wL³/6EI.
- Peak at free end: θ = wL³/8EI, δ = 11wL⁴/120EI.
Practice more: this is question 40 in Slope and Deflection MCQs – Part 3. Try the whole set and check every answer instantly.