A cantilever of span L carries a uniformly distributed load of w per unit length. The maximum deflection in the beam is (E = modulus of elasticity, I = moment of inertia):
Try it first: pick an option, then read the explanation.
Correct answer: A. wL⁴/(8EI)
Asked in SSC JE 23/03/2021 (Evening); also 24/01/2018 (Evening)
Detailed Explanation
For a UDL on a cantilever, M = −w(L − x)²/2. Integrating twice with zero slope and deflection at the fixed end gives the free-end values θ = wL³/6EI and δ = wL⁴/8EI.
\[ EIy = -\frac{w}{24}\left(x^4 - 4Lx^3 + 6L^2x^2\right) \Rightarrow y(L) = -\frac{w}{24}(3L^4) = -\frac{wL^4}{8EI} \]
Why “wL⁴/(8EI)” Is Correct
The maximum deflection, at the free end, is wL⁴/(8EI).
Why the Other Options Are Wrong
- B. wL²/(6EI): wL²/6EI has wrong dimensions (the slope is wL³/6EI).
- C. wL⁴/(48EI): wL⁴/48EI is not a standard result for this case.
- D. wL²/(8EI): wL²/8EI is the form of a bending moment (wL²/8) divided by EI — dimensionally not a deflection.
Key Point
Cantilever with UDL: θ = wL³/6EI, δ = wL⁴/8EI.
Key Concepts for Students
- With total load W = wL, δ = WL³/8EI.
- Simply supported UDL: 5wL⁴/384EI — about 1/48 of the cantilever value.
- Deflection is proportional to L⁴ for distributed loads.
Practice more: this is question 15 in Slope and Deflection MCQs – Part 1. Try the whole set and check every answer instantly.