Beams and Slabs
JKSSB Junior Engineer (Civil) · RCC Structures · 24 practice questions with answers and explanations: 14 Easy, 9 Medium, 1 Hard.
Sample questions
Sample Easy question 1
Compression steel in a doubly reinforced beam also helps to:
- Reduce the tension steel needed to zero
- Increase ductility and reduce the long-term deflection
- Increase the cover
- Resist shear alone
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Answer: B. It reduces the effect of creep on the deflection.
Sample Easy question 2
The minimum area of tension reinforcement in a beam is:
- 0.12% of bD
- 0.4 ÷ fy of bd
- 0.85 ÷ fy of bd
- 1% of bD
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Answer: C. As,min ÷ bd = 0.85 ÷ fy, with fy in N/mm².
Sample Easy question 3
The maximum area of tension reinforcement in a beam is:
- 1% of the gross area
- 4% of the gross cross-sectional area
- 6% of the gross area
- 2% of the gross area
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Answer: B. IS 456 limits it to 0.04 bD.
Sample Medium question 4
The minimum steel in a 150 mm thick slab using high-strength deformed bars, per metre width, is:
- 225 mm²
- 360 mm²
- 120 mm²
- 180 mm²
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Answer: D. 0.12% of 150 × 1,000 = 180 mm².
Sample Medium question 5
For a beam with an effective depth of 360 mm, the maximum spacing of vertical stirrups (the smaller of 0.75 d and 300 mm) is:
- 270 mm
- 300 mm
- 540 mm
- 360 mm
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Answer: A. 0.75 × 360 = 270 mm, which is less than 300 mm.
Sample Medium question 6
A simply supported beam has a clear span of 4.0 m, an effective depth of 0.4 m and supports 0.3 m wide. The effective span is the smaller of the centre-to-centre of supports and clear span plus d, which is:
- 4.3 m
- 4.4 m
- 4.6 m
- 4 m
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Answer: A. Centre to centre = 4.0 + 0.3 = 4.3 m, and clear + d = 4.4 m. The smaller is 4.3 m.
Sample Hard question 7
For a T-beam with a span lo = 6,000 mm, web width 300 mm and flange thickness 120 mm, the effective flange width of a continuous T-beam (bf = lo ÷ 6 + bw + 6 Df) is about:
- 1,000 mm
- 2,020 mm
- 1,020 mm
- 4,040 mm
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Answer: B. bf = 6,000 ÷ 6 + 300 + 6 × 120 = 2,020 mm.