Limit State Design
JKSSB Junior Engineer (Civil) · RCC Structures · 21 practice questions with answers and explanations: 11 Easy, 8 Medium, 2 Hard.
Sample questions
Sample Easy question 1
The partial load factor for dead load plus live load in the limit state of collapse is:
- 1.5
- 1.2
- 1.0
- 2.0
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Answer: A. IS 456 uses 1.5 (DL + LL).
Sample Easy question 2
In the rectangular stress block, the compressive force in the concrete is:
- 0.87 fck b xu
- 0.45 fck b xu
- 0.36 fck b xu
- 0.67 fck b xu
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Answer: C. The average stress is 0.36 fck over the depth xu, acting at 0.416 xu.
Sample Easy question 3
The code limits the width of cracks in a member exposed to moderate conditions to about:
- 0.03 mm
- 0.3 mm
- 3 mm
- 5 mm
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Answer: B. This is a serviceability limit state.
Sample Medium question 4
The span to effective depth ratio of a simply supported beam, as a basic value in IS 456, is:
- 20
- 26
- 40
- 7
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Answer: A. It is 7 for a cantilever and 26 for a continuous beam.
Sample Medium question 5
A beam carries a dead load of 20 kN/m and a live load of 10 kN/m. The design (factored) load is:
- 60 kN/m
- 30 kN/m
- 36 kN/m
- 45 kN/m
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Answer: D. wu = 1.5 (DL + LL) = 1.5 × 30 = 45 kN/m.
Sample Medium question 6
The area of steel in four bars of 20 mm diameter is about:
- 1,257 mm²
- 628 mm²
- 251 mm²
- 2,513 mm²
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Answer: A. A = 4 × (π ÷ 4) × 20² = 1,257 mm².
Sample Hard question 7
A beam 230 mm wide has Ast = 942 mm² of Fe 415 steel in M20. The depth of the neutral axis xu = 0.87 fy Ast ÷ (0.36 fck b) is about:
- 103 mm
- 205 mm
- 225 mm
- 411 mm
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Answer: B. xu = 0.87 × 415 × 942 ÷ (0.36 × 20 × 230) = 205 mm.
Sample Hard question 8
The limiting moment of resistance of a 300 mm wide beam with an effective depth of 460 mm in M25 concrete and Fe 415 steel is about:
- 109 kN·m
- 438 kN·m
- 62 kN·m
- 219 kN·m
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Answer: D. Mu,lim = 0.138 fck b d² = 0.138 × 25 × 300 × 460² = 219 kN·m.