Concrete structures miscellaneous

Concrete structures miscellaneous

Easy Questions

Concrete Structures

Concrete structures miscellaneous
  1. The flexural tensile strength of M25 grade of concrete, in N/mm2, as per IS: 456-2000 is _______








  1. View Hint View Answer Discuss in Forum

    Flexural tensile strength = 0.7 √ fck
    = 0.7 × √25 = 3.5 N/mm2.

    Correct Option: C

    Flexural tensile strength = 0.7 √ fck
    = 0.7 × √25 = 3.5 N/mm2.

  1. The development length of a deformed reinforcement bar can be expressed as (1/k) (Φσs/ τbd).
    From the IS:456-2000, the value of k can be calculated as _________.








  1. View Hint View Answer Discuss in Forum

    Ld =
    Φσst
    4τbd

    For deformed bars, τbd is increased by 60%
    ∴ Ld =
    Φσst
    4× (τbd × 1.6)

    =
    Φσst
    6.4τbd

    =
    Φσst
    kτbd

    ∴ k = 6.4

    Correct Option: A

    Ld =
    Φσst
    4τbd

    For deformed bars, τbd is increased by 60%
    ∴ Ld =
    Φσst
    4× (τbd × 1.6)

    =
    Φσst
    6.4τbd

    =
    Φσst
    kτbd

    ∴ k = 6.4

  1. The composition of an air-entrained concrete is given below:
    Water: 184 kg/m2
    Ordinary Portland Cement(OPC): 368 kg/m3
    Sand: 606 kg/m3
    Coarse aggregate: 1155 kg/m3
    Assume the specific gravity of OPC, sand and coarse aggregate to be 3.14, 2.67 and 2.74, respectively. The air content is ______ litres/m3.








  1. View Hint View Answer Discuss in Forum

    Mc
    		+
    Ms
    		+
    Ma
    		+ Vw + Va = 1
    ρcρsρa

    ⇒
    368
    		+
    606
    3.14 × 10002.67 × 1000

    +
    1155
    		+
    184
    		+ Va = 1
    2.74 × 10001000

    ⇒ 0.117 + 0.227 + 0.421 + 0.184 + Va = 1
    ∴Va = 0.051 = 51 l/m3

    Correct Option: A

    Mc
    		+
    Ms
    		+
    Ma
    		+ Vw + Va = 1
    ρcρsρa

    ⇒
    368
    		+
    606
    3.14 × 10002.67 × 1000

    +
    1155
    		+
    184
    		+ Va = 1
    2.74 × 10001000

    ⇒ 0.117 + 0.227 + 0.421 + 0.184 + Va = 1
    ∴Va = 0.051 = 51 l/m3

  1. Consider the singly reinforced beam section given below(left figure). the stress block parameters for the cross- section from IS: 456-2000 are also given below (right figure). The moment of resistance for the given section by the limit state method is ______ kN-m.








  1. View Hint View Answer Discuss in Forum

    Ast = 4 ×
    π
    		d2
    4

    = 4 ×
    π
    		(12)2
    4

    = 453 mm2
    0.36 fck. b. xu = 0.87 fy. Ast
    ⇒ xu =
    0.87fy.Ast
    0.36 fck .b

    =
    0.87 × 415 × 453
    		= 90.86 mm
    0.36 × 25 × 200

    xu. max = 0.48d
    = 0.48 × 300 = 144 mm
    ∴ xu < xu max
    ∴ it is U.R section (under reinforced)
    Moment of resistance,
    MV = 0.87 × fy × Ast × (d – 0.42 xu)
    = 0.87 × 415 × 453 (300 – 0.42 × 90.86) = 42.82 kNm

    Correct Option: C

    Ast = 4 ×
    π
    		d2
    4

    = 4 ×
    π
    		(12)2
    4

    = 453 mm2
    0.36 fck. b. xu = 0.87 fy. Ast
    ⇒ xu =
    0.87fy.Ast
    0.36 fck .b

    =
    0.87 × 415 × 453
    		= 90.86 mm
    0.36 × 25 × 200

    xu. max = 0.48d
    = 0.48 × 300 = 144 mm
    ∴ xu < xu max
    ∴ it is U.R section (under reinforced)
    Moment of resistance,
    MV = 0.87 × fy × Ast × (d – 0.42 xu)
    = 0.87 × 415 × 453 (300 – 0.42 × 90.86) = 42.82 kNm

  1. A simply supported reinforced concrete beam of length 10 m sags while undergoing shrinkage. Assuming a uniform curvature of 0.004 m-1 along the span, the maximum deflection (in m) of the beam at mid-span is ____________








  1. View Hint View Answer Discuss in Forum

    R =
    1
    		+
    1
    		= 250 m
    ψ0.04

    Deflected length

    = 249.9995 m
    ∴ Deflection = 250 – 249.9995 = 0.0005 m

    Correct Option: A

    R =
    1
    		+
    1
    		= 250 m
    ψ0.04

    Deflected length

    = 249.9995 m
    ∴ Deflection = 250 – 249.9995 = 0.0005 m