A beam of span 5 m, fixed at A and B, carries a point load of 50 kN at 2 m from A. The fixed-end moments at supports A and B respectively are:
Try it first: pick an option, then read the explanation.
Asked in SSC JE 30/10/2020 (Morning)
Detailed Explanation
The English text of option (b) printed "36 kN·m clockwise" for B; the Hindi version and the calculation give 24 kN·m, so option (b) is shown as "36 kN·m anticlockwise and 24 kN·m clockwise".
For a point load W at a from A and b from B, the fixed-end moments are MA = Wab²/L² and MB = Wa²b/L². The larger moment is at the support nearer the load.
\[ M_A = \frac{50\times2\times3^2}{5^2} = 36\ \text{kN·m}, \qquad M_B = \frac{50\times2^2\times3}{5^2} = 24\ \text{kN·m} \]
Why “36 kN·m anticlockwise and 24 kN·m clockwise” Is Correct
Both end moments are hogging. Acting on the beam, the wall at A applies an anticlockwise 36 kN·m and the wall at B a clockwise 24 kN·m (they resist the beam ends rotating downward towards the load). That is option (b).
Why the Other Options Are Wrong
- A. 24 kN·m clockwise and 24 kN·m clockwise: Neither magnitude nor sense is right: the end moments are unequal (36 and 24).
- C. 24 kN·m anticlockwise and 36 kN·m anticlockwise: Swaps the magnitudes — the bigger moment is at A, nearer the load.
- D. 36 kN·m clockwise and 24 kN·m anticlockwise: Right magnitudes but reversed senses.
Key Point
FEM: MA = Wab²/L², MB = Wa²b/L²; larger moment at the end nearer the load.
Key Concepts for Students
- Reactions: RA = 32.4 kN, RB = 17.6 kN.
- Sagging under the load = 2Wa²b²/L³ = 28.8 kN·m.
- Option (b) corrected to match the Hindi text.
Practice more: this is question 2 in Fixed Beam MCQs – Part 1. Try the whole set and check every answer instantly.