The slope at the supports of a simply supported beam of effective span L carrying a central point load W is:
Try it first: pick an option, then read the explanation.
Correct answer: C. WL²/(16EI)
Asked in SSC JE 2012
Detailed Explanation
By symmetry the slope at mid-span is zero, so each half acts like a cantilever of length L/2 fixed at the centre, loaded by the support reaction W/2 at its tip.
\[ \theta_A = \frac{(W/2)(L/2)^2}{2EI} = \frac{WL^2}{16EI} \]
Moment-area check: half the BMD area = \( \tfrac12\cdot\tfrac{L}{2}\cdot\tfrac{WL}{4} = \tfrac{WL^2}{16} \).
Why “WL²/(16EI)” Is Correct
The support slope is WL²/(16EI).
Why the Other Options Are Wrong
- A. WL²/(8EI): WL²/8EI is twice the correct value.
- B. WL²/(12EI): WL²/12EI is not a standard result.
- D. WL²/(24EI): WL²/24EI resembles the UDL slope wL³/24EI with the wrong load term.
Key Point
Simply supported, central W: θ = WL²/16EI, δ = WL³/48EI.
Key Concepts for Students
- Mohr I: slope at support = area of M/EI between support and centre.
- UDL: θ = wL³/24EI.
- Equal end couples: θ = ML/2EI.
Practice more: this is question 38 in Slope and Deflection MCQs – Part 3. Try the whole set and check every answer instantly.