A cantilever 2 m long and 10 cm deep tapers in plan from a width of 24 cm at the fixed end to zero at its free end, and carries a point load of 5 kN at the free end. If E = 0.20 × 10⁶ N/mm², the deflection of the free end is:
Try it first: pick an option, then read the explanation.
Asked in SSC JE 04/03/2017 (Evening)
Detailed Explanation
Note on this question: the exam omitted the load; the book's key is 5 mm, which corresponds to a 5 kN point load at the free end, so that load is stated here.
Because the width grows linearly from the tip (b = b₀x/L) and the moment from a tip load also grows linearly (M = Wx), the ratio M/I is the same at every section. The beam bends into a circular arc of constant curvature — it is a beam of uniform strength.
\[ I_0 = \frac{b_0 d^3}{12} = \frac{240\times100^3}{12} = 2\times10^{7}\ \text{mm}^4 \qquad \frac{1}{R} = \frac{M}{EI} = \frac{WL}{EI_0} = \frac{5000\times2000}{2\times10^5\times2\times10^7} = 2.5\times10^{-6}\ \text{mm}^{-1} \]
\[ \delta = \frac{L^2}{2R} = \frac{2000^2\times2.5\times10^{-6}}{2} = 5\ \text{mm} \quad\left(\text{general: } \delta = \frac{6WL^3}{Eb_0d^3}\right) \]
Why “5 mm” Is Correct
With constant curvature, δ = L²/2R, which is 1.5 times the deflection of a prismatic cantilever of width b₀ (WL³/3EI₀). The free-end deflection is 5 mm.
Why the Other Options Are Wrong
- A. 2 mm: 2 mm would need a 2 kN load.
- B. 3 mm: 3 mm would need a 3 kN load.
- C. 4 mm: 4 mm would need a 4 kN load (or the prismatic formula with a different load).
Key Point
Plan-tapered cantilever (b ∝ x) with tip load: uniform curvature, δ = WL³/(2EI₀) = 6WL³/(Eb₀d³).
Key Concepts for Students
- Prismatic cantilever of the root section: δ = WL³/3EI₀ — the tapered one is 1.5× more flexible.
- Constant M/I ⇒ constant bending stress ⇒ uniform strength.
- Leaf springs use this principle.
Practice more: this is question 31 in Slope and Deflection MCQs – Part 3. Try the whole set and check every answer instantly.