Concrete structures miscellaneous
- According to the concept of Limit State Design as per IS 456:2000, the probability of failure of a structure is __________
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As per IS 456:2000
Correct Option: A
As per IS 456:2000
- A rectangular concrete beam of width 120 mm and depth 200 mm is pre-stressed by pre-tensioning to a force of 150 kN at an eccentricity of 20 mm. The cross sectional area of the prestressing steel is 187.5 mm2. Take modulus of elasticity of steel and concrete as 2.1 × 105 MPa and 3.0 × 104 MPa respectively. The percentage loss of stress in the prestressing steel due to elastic deformation of concrete is
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Initial direct stress
= P = 150 × 103 A bd = 150 × 103 = 6.25 N/mm2 120 × 200 m = f I y ∴ f = Pe × e I = 150 × 103 × 20 × 20 1 × 120 × (200)3 12
= 0.75 N/mm2
Max. compression in concrete = 6.25 + 0.75
= 7 N/mm2
strain in concrete= fs = 7 Es 3 × 104
= 2.33 ×10-4
loss of tensile stress = 2.33 × 10-4 × 2.1 × 105 = 40 N/mm2Initial stress = 150 × 103 = 800 N/mm2 187.5 % loss = 49 = 6.125% 800 Correct Option: B
Initial direct stress
= P = 150 × 103 A bd = 150 × 103 = 6.25 N/mm2 120 × 200 m = f I y ∴ f = Pe × e I = 150 × 103 × 20 × 20 1 × 120 × (200)3 12
= 0.75 N/mm2
Max. compression in concrete = 6.25 + 0.75
= 7 N/mm2
strain in concrete= fs = 7 Es 3 × 104
= 2.33 ×10-4
loss of tensile stress = 2.33 × 10-4 × 2.1 × 105 = 40 N/mm2Initial stress = 150 × 103 = 800 N/mm2 187.5 % loss = 49 = 6.125% 800
- A concrete beam pre-stressed with a parabolic tendon is shown in the sketch. The eccentricity of the tendon is measured from the centroid of the cross-section. The applied pre-stressing force at service is 1620 kN. The uniformly distributed load of 45 kN/m includes the self-weight.
The stress (in N/mm2) in the bottom fibre at mid-span is
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Total stress bottom,σb = P + Pe - m A Zb Zb I = bd3 12 
= 500 × 7502 = 4687500 mm2 6
Zb = Zt.
= 2.93Correct Option: B
Total stress bottom,σb = P + Pe - m A Zb Zb I = bd3 12 = 500 × 7502 = 4687500 mm2 6
Zb = Zt.
= 2.93
- Maximum possible value of Compacting Factor for fresh (green) concrete is:
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0.5
Correct Option: A
0.5
- 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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Factored SF, Vu = 45 kN
τv = Vu = 45 × 103 bd 230 × 450
= 0.4348 N/mm2% tensile steel = A bd 
= 0. 437%τc = 0.36 + 0.12 (0.437 - 0.25) 0.25
= 0.45 N/mm2
∵ τv – τc < 0,
minimum shear reinforcement is given by:Asv = 0.4 bSv 0.87 × fy Asv = 0.4 × Sv × b 0.87 × fy
We limit fy to 415 N/mm2∴ Asv = 2 × π × Φ2 4 = 0.4 × 325 × 230 = 82.814 mm2 0.87 × 415
Φ = 7.26 mm
≈ 8 mmCorrect Option: A
Factored SF, Vu = 45 kN
τv = Vu = 45 × 103 bd 230 × 450
= 0.4348 N/mm2% tensile steel = A bd
= 0. 437%τc = 0.36 + 0.12 (0.437 - 0.25) 0.25
= 0.45 N/mm2
∵ τv – τc < 0,
minimum shear reinforcement is given by:Asv = 0.4 bSv 0.87 × fy Asv = 0.4 × Sv × b 0.87 × fy
We limit fy to 415 N/mm2∴ Asv = 2 × π × Φ2 4 = 0.4 × 325 × 230 = 82.814 mm2 0.87 × 415
Φ = 7.26 mm
≈ 8 mm
