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A rectangular beam of width (b) 230 mm and effective depth (d) 450 mm is reinforced with four bars of 12 mm diameter. The grade of concrete is M20 and grade of steel is Fe500. Given that for M20 grade of concrete the ultimate shear strength, τuc = 0.36 N/mm2 for steel percentage, p = 0.25, and τuc = 0.48 N/mm2 for p = 0.50. For a factored shear force of 45 kN, the diameter (in mm) of Fe500 steel two lagged stirrups to be used at spacing of 375 mm, should be
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Correct Option: A
Factored SF, Vu = 45 kN
| τv = | = | |||
| bd | 230 × 450 |
= 0.4348 N/mm2
| % tensile steel = | bd |
= 0. 437%
| τc = 0.36 + | (0.437 - 0.25) | 0.25 |
= 0.45 N/mm2
∵ τv – τc < 0,
minimum shear reinforcement is given by:
| = | ||||
| bSv | 0.87 × fy |
| Asv = | 0.87 × fy |
We limit fy to 415 N/mm2
| ∴ Asv = 2 × | × Φ2 | 4 |
| = | = 82.814 mm2 | 0.87 × 415 |
Φ = 7.26 mm
≈ 8 mm
