A cantilever of length L carries a uniformly varying load that is zero at the free end and W kN/m at the fixed end. The deflection at the free end is:
Try it first: pick an option, then read the explanation.
Correct answer: D. WL⁴/(30EI)
Asked in SSC JE 30/10/2020 (Evening)
Detailed Explanation
The load intensity grows linearly from zero at the free end to W at the support. With the peak near the support, the load has a small lever arm, so the deflection is much smaller than for a UDL (wL⁴/8EI).
\[ M(x) = -\frac{W(L-x)^3}{6L},\quad \text{integrate twice with } y(0) = y'(0) = 0 \Rightarrow y(L) = -\frac{WL^4}{30EI} \]
Why “WL⁴/(30EI)” Is Correct
The free-end deflection is WL⁴/(30EI), with free-end slope WL³/24EI.
Why the Other Options Are Wrong
- A. WL⁴/(385EI): 385 resembles the 384 of the simply supported UDL formula — a mix-up.
- B. WL⁴/(45EI): WL⁴/45EI is not a standard cantilever result.
- C. WL⁴/(60EI): WL⁴/60EI is half the correct value.
Key Point
Cantilever, triangular load (max at support): δ = wL⁴/30EI, θ = wL³/24EI.
Key Concepts for Students
- Peak at the free end instead: δ = 11wL⁴/120EI.
- UDL: wL⁴/8EI.
- Total triangular load = wL/2 acting at L/3 from the fixed end.
Practice more: this is question 16 in Slope and Deflection MCQs – Part 2. Try the whole set and check every answer instantly.