Question 1 |

M20 concrete as per IS 456: 2000 refers to concrete with a design mix having

an average cube strength of 20 \mathrm{MPa} | |

an average cylinder strength of 20 \mathrm{MPa} | |

a 5-percentile cube strength of 20 \mathrm{MPa} | |

a 5-percentile cylinder strength of 20 \mathrm{MPa} |

Question 1 Explanation:

In M20, M refers to mix and 20 to characteristic cube strength. As per clause no. 6.1.1, IS456: 2000 characteristic strength is defined as the strength below which not more than 5 percent of the test results are expected to fall.

Hence, correct option is (C).

Hence, correct option is (C).

Question 2 |

A reinforced concrete beam with rectangular cross section (width = 300 mm,
effective depth = 580 mm) is made of M30 grade concrete. It has 1%
longitudinal tension reinforcement of Fe 415 grade steel. The design shear
strength for this beam is 0.66 N/mm^2. The beam has to resist a factored shear
force of 440 kN. The spacing of two-legged, 10 mm diameter vertical stirrups of
Fe 415 grade steel is ______mm. (round off to the nearest integer)

127 | |

101 | |

254 | |

331 |

Question 2 Explanation:

b=300mm

d=580 mm

V_u=440 kN

Concrete used is M30

Raft steel is Fe415

V_{cu}=\tau _c Bd=0.66 \times 300 \times \frac{580}{1000}=114.84 kN

V_{su}=V_u-V_{cu}=440-114.84=325.16 kN

Spacing of 2-legged shear reinforcement

\begin{aligned} s_V&=\frac{A_{SV} \times 0.87 f_y \times d}{V_{su}}\\ &=\frac{2 \times \frac{\pi}{2} \times (10)^2 \times 0.87 \times 415 \times 580}{325.16 \times 1000}\\ &=101.16 mm \end{aligned}

d=580 mm

V_u=440 kN

Concrete used is M30

Raft steel is Fe415

V_{cu}=\tau _c Bd=0.66 \times 300 \times \frac{580}{1000}=114.84 kN

V_{su}=V_u-V_{cu}=440-114.84=325.16 kN

Spacing of 2-legged shear reinforcement

\begin{aligned} s_V&=\frac{A_{SV} \times 0.87 f_y \times d}{V_{su}}\\ &=\frac{2 \times \frac{\pi}{2} \times (10)^2 \times 0.87 \times 415 \times 580}{325.16 \times 1000}\\ &=101.16 mm \end{aligned}

Question 3 |

A reinforcing steel bar, partially embedded in concrete, is subjected to a tensile force P. The figure that appropriately represents the distribution of the magnitude of bond stress
(represented as hatched region), along the embedded length of the bar, is

A | |

B | |

C | |

D |

Question 4 |

Tie bars of 12 mm diameter are to be provided in a concrete pavement slab. The working tensile stress of the tie bars is 230 MPa, the average bond strength between a tie bar and concrete is 2 MPa, and the joint gap between the slabs is 10 mm. Ignoring the loss of bond and the tolerance factor, the design length of the tie bars (in mm, round off to the nearest integer) is _____

350 | |

700 | |

1050 | |

850 |

Question 4 Explanation:

L_d=\frac{\phi \sigma _{st}}{4\tau _{bd}}=\frac{12 \times 230}{4 \times 2}=345mm

Length of tie bar =L_d+10+L_d=345+10+345=700mm

Question 5 |

For a given loading on a rectangular plain concrete beam with an overall depth of 500 mm, the compressive strain and tensile strain developed at the extreme fibers are of the same magnitude of 2.5 \times 10^{-4}. The curvature in the beam cross-section (in m^{-1}, round off to 3 decimal places), is ____

0.008 | |

0.006 | |

0.004 | |

0.001 |

Question 5 Explanation:

Simple bending equation

\begin{aligned} \frac{m}{I}&=\frac{f}{y}=\frac{E}{R} \\ \frac{1}{R}&=\frac{f}{E}\cdot \frac{1}{y} \\ &=\frac{\varepsilon }{y}=\frac{2.5 \times 10^{-4}}{(0.5/2)}=0.001m^{-1} \end{aligned}

\begin{aligned} \frac{m}{I}&=\frac{f}{y}=\frac{E}{R} \\ \frac{1}{R}&=\frac{f}{E}\cdot \frac{1}{y} \\ &=\frac{\varepsilon }{y}=\frac{2.5 \times 10^{-4}}{(0.5/2)}=0.001m^{-1} \end{aligned}

There are 5 questions to complete.