A cantilever carrying a point load of 10 units at its free end has a maximum deflection of 1250/(3EI) units. The slope at its free end is:
Try it first: pick an option, then read the explanation.
Asked in SSC JE 15/11/2022 (Shift III)
Detailed Explanation
Use the deflection formula to find the length, then use the slope formula. Both are standard results for a cantilever with a tip load.
\[ \frac{Pl^3}{3EI} = \frac{1250}{3EI} \Rightarrow 10\,l^3 = 1250 \Rightarrow l = 5 \]
\[ \theta = \frac{Pl^2}{2EI} = \frac{10\times25}{2EI} = \frac{125}{EI} \]
Shortcut: \(\theta = \dfrac{3\delta}{2l} = \dfrac{3}{2\times5}\times\dfrac{1250}{3EI} = \dfrac{125}{EI}\).
Why “125/(EI) units” Is Correct
The length works out to 5 units and the free-end slope to 125/EI units.
Why the Other Options Are Wrong
- A. 250/(EI) units: 250/EI is twice the correct value (Pl²/EI).
- B. 62.5/(EI) units: 62.5/EI is half the correct value (Pl²/4EI).
- D. 12.5/(EI) units: 12.5/EI uses l instead of l².
Key Point
Cantilever with tip load: θ = Pl²/2EI and δ = Pl³/3EI, so δ = (2/3)θl.
Key Concepts for Students
- Ratio δ/θ = 2l/3 for a tip-loaded cantilever.
- For a UDL cantilever, δ/θ = 3l/4.
- Always solve for the unknown dimension first.
Practice more: this is question 11 in Slope and Deflection MCQs – Part 1. Try the whole set and check every answer instantly.