Two simply supported beams carry central concentrated loads. Beam A: length 5 m, rigidity EI, load 2 kN. Beam B: length 10 m, rigidity 2EI, load 1 kN. Compare the slopes at the ends.
Try it first: pick an option, then read the explanation.
Asked in SSC JE 05/06/2024 (9 am–11 am)
Detailed Explanation
This slope question is placed in this chapter by the book. Disputed key: the book notes that the commission's key marked (d), but the standard formula shows the two slopes are exactly equal.
For a simply supported beam with a central load W, the slope at each support is θ = WL²/(16EI).
\[ \theta_A = \frac{2\times5^2}{16\,EI} = \frac{50}{16EI} = \frac{25}{8EI} \qquad \theta_B = \frac{1\times10^2}{16\times2EI} = \frac{100}{32EI} = \frac{25}{8EI} \]
Why “Slope of beam A = slope of beam B” Is Correct
Both work out to 25/(8EI), so the slopes are equal. Beam B is twice as long (×4 in L²) but has half the load and twice the rigidity (÷4), which cancel exactly. We give (a) and flag the reported key (d).
Why the Other Options Are Wrong
- B. Slope of beam A > slope of beam B: θA is not larger — both equal 25/8EI.
- C. Insufficient data to compare the slopes: All data needed (W, L, EI) are given.
- D. Slope of beam A < slope of beam B: θA is not smaller — the reported key (d) does not match the formula.
Key Point
SS central load: θ = WL²/16EI. Here W·L²/EI is the same for both beams, so the slopes are equal.
Key Concepts for Students
- Deflections would differ: δ = WL³/48EI gives δB = 2δA.
- Compare by the combination W L²/EI rather than each value separately.
- Reported key (d) is flagged as disputed on this page.
Practice more: this is question 1 in Principal Stress and Principal Planes MCQs – Part 1. Try the whole set and check every answer instantly.