A cantilever 2 m long, fixed at its left end, carries a downward point load of 9 kN at its right (free) end. Find the magnitude and direction of the couple to be applied at the free end so that the vertical deflection there is zero.
Try it first: pick an option, then read the explanation.
Asked in SSC JE 02/03/2017 (Morning)
Detailed Explanation
Disputed key: the book's key is (c) 12 kN·m clockwise. The magnitude is right, but a clockwise couple at the right-hand free end would bend the tip further down. To lift the tip, the couple must produce sagging, which at the right end of a beam is an anticlockwise couple.
\[ \frac{WL^3}{3EI} = \frac{ML^2}{2EI} \Rightarrow M = \frac{2WL}{3} = \frac{2\times9\times2}{3} = 12\ \text{kN·m} \]
Why “12 kN·m anticlockwise” Is Correct
The 9 kN load causes hogging (tip goes down by WL³/3EI). An anticlockwise couple at the free end causes uniform sagging and lifts the tip by ML²/2EI. Equating the two gives 12 kN·m anticlockwise. We give (d) and flag the book's (c).
Why the Other Options Are Wrong
- A. 9 kN·m clockwise: 9 kN·m (= WL/2) would cancel only 3/4 of the deflection, and clockwise would add to it.
- B. 9 kN·m anticlockwise: 9 kN·m is the right direction but too small.
- C. 12 kN·m clockwise: Correct magnitude, wrong direction (book's key): a clockwise couple at the right end increases the downward deflection.
Key Point
Zero tip deflection: M = 2WL/3, applied so that it causes sagging (anticlockwise at a right-hand free end).
Key Concepts for Students
- Couple at free end: δ = ML²/2EI, θ = ML/EI.
- Zero tip slope instead would need M = WL/2.
- Book key (c) is flagged as disputed on this page.
Practice more: this is question 33 in Slope and Deflection MCQs – Part 3. Try the whole set and check every answer instantly.