Concrete structures miscellaneous
- Maximum strain in an extreme fibre in concrete and in the tension reinforcement (Fe-415 grade and Es = 200 kN/mm2) in a balanced section at limit state of flexure are respectively
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Max strain in steel
= 0.002 + 0.87 fy Es = 0.002 + 0.87 × 415 = 0.0038 200 × 103
For concrete max strain = 0.0035Correct Option: A
Max strain in steel
= 0.002 + 0.87 fy Es = 0.002 + 0.87 × 415 = 0.0038 200 × 103
For concrete max strain = 0.0035
- As per the provisions of IS 456-2000, the (short-term) modulus of elasticity of M 25 grade concrete (in N/mm2) can be assumed to be
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EC = 5000 √25
= 25000 MPaCorrect Option: A
EC = 5000 √25
= 25000 MPa
- As per provisions of IS 456-2000, in the limit state method for design of beams, the limiting value of the depth of neutral axis in a reinforced concrete beam of effective depth ‘d’ is given as
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Limiting value of depth of neutral axis depends on grade of concrete.
Correct Option: D
Limiting value of depth of neutral axis depends on grade of concrete.
- Match the information related to tests on aggregates given in Group-I with that in Group-II
Group-I
P. Resistance to impact
Q. Resistance to wear
R. Resistance to weathering action
S. Resistance to crushing
Group-II
1. Hardness
2. Strength
3. Toughness
4. Soundness
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Resistance to impact – Toughness
Resistance to wear – Hardness
Resistance to weathering action – Soundness
Resistance to crushing – StrengthCorrect Option: B
Resistance to impact – Toughness
Resistance to wear – Hardness
Resistance to weathering action – Soundness
Resistance to crushing – Strength
- Top ring beam of an Intze tank carries a hoop tension of 120 kN. the beam cross-section is 250 mm wide the 400 mm deep and it is reinforced with 4 bars of 20 mm diameter of Fe 415 grade. Modular ratio of the concrete is 10. The tensile stress in N/ mm2 in the concrete is
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Area of reinforcement
= 4 × π × 202 = 1256.63 mm2 4
Equivalent area of concrete = m. Ast + (250 × 400 – Ast)
= 10 × 1256.6 + (250 × 400 – 1256.6)
= 111309.6 mm2
Stress in concrete= 120 × 103 = 1.078 N/mm2 11309.6 Correct Option: B
Area of reinforcement
= 4 × π × 202 = 1256.63 mm2 4
Equivalent area of concrete = m. Ast + (250 × 400 – Ast)
= 10 × 1256.6 + (250 × 400 – 1256.6)
= 111309.6 mm2
Stress in concrete= 120 × 103 = 1.078 N/mm2 11309.6