Question 1 |

Consider the singly reinforced section of a cantilever concrete beam under bending, as shown in the figure (M25 grade concrete, Fe415 grade steel). The stress block parameters for the section at ultimate limit state, as per IS 456: 2000 notations, are given. The ultimate moment of resistance for the section by the Limit State Method is ____ kN.m (round off to one decimal place).

[Note: Here, As is the total area of tension steel bars, b is the width of the section, d is the effective depth of the bars, f_{c k} is the characteristic compressive cube strength of concrete, f_{y} is the yield stress of steel, and \mathrm{xu} is the depth of neutral axis.]

[Note: Here, As is the total area of tension steel bars, b is the width of the section, d is the effective depth of the bars, f_{c k} is the characteristic compressive cube strength of concrete, f_{y} is the yield stress of steel, and \mathrm{xu} is the depth of neutral axis.]

301 | |

225.2 | |

365.4 | |

154.8 |

Question 1 Explanation:

Given data,

Grade of concrete = M25

Grade of Steel =\mathrm{Fe} 415

Effective depth, d=600-45=555 \mathrm{~mm}

A_{s}=3 \times \frac{\pi}{4} \times(28)^{2}=1847.26 \mathrm{~mm}^{2}

From Compression = Tension

Depth of neutral axis, x_{u}=\frac{0.87 f_{y} A_{s}}{0.36 f_{c k} b}=\frac{0.87 \times 415 \times 1847.26}{0.36 \times 25 \times 300} =247.02 \mathrm{~mm}

Limiting depth of neutral axis, x_{u, l i m}=0.48 \mathrm{~d} =0.48 \times 555=266.40 \mathrm{~mm}

\Rightarrow Section is under reinforced.

Ultimate moment of resistance

=0.36 f_{c k} b x_{u}\left(d-0.42 x_{u}\right)

=0.36 \times 25 \times 300 \times 247.02 \times (555-0.42 \times 247.02) \times 10^{-6}

=301.0 (after rounding off to one decimal place).

Grade of concrete = M25

Grade of Steel =\mathrm{Fe} 415

Effective depth, d=600-45=555 \mathrm{~mm}

A_{s}=3 \times \frac{\pi}{4} \times(28)^{2}=1847.26 \mathrm{~mm}^{2}

From Compression = Tension

Depth of neutral axis, x_{u}=\frac{0.87 f_{y} A_{s}}{0.36 f_{c k} b}=\frac{0.87 \times 415 \times 1847.26}{0.36 \times 25 \times 300} =247.02 \mathrm{~mm}

Limiting depth of neutral axis, x_{u, l i m}=0.48 \mathrm{~d} =0.48 \times 555=266.40 \mathrm{~mm}

\Rightarrow Section is under reinforced.

Ultimate moment of resistance

=0.36 f_{c k} b x_{u}\left(d-0.42 x_{u}\right)

=0.36 \times 25 \times 300 \times 247.02 \times (555-0.42 \times 247.02) \times 10^{-6}

=301.0 (after rounding off to one decimal place).

Question 2 |

Two plates are connected by fillet welds of size 10 \mathrm{mm} and subjected to tension, as shown in the figure. The thickness of each plate is 12 \mathrm{~mm}. The yield stress and the ultimate stress of steel under tension are 250 \mathrm{MPa} and 410 \mathrm{MPa}, respectively. The welding is done in the workshop (partial safety factor, \gamma_{\mathrm{mw}}=1.25 ). As per the Limit State Method of IS 800: 2007, what is the minimum length (in \mathrm{mm}, rounded off to the nearest higher multiple of 5 \mathrm{~mm} ) required of each weld to transmit a factored force P equal to 275 \mathrm{kN} ?

100 | |

105 | |

110 | |

115 |

Question 2 Explanation:

Size of weld =10 \mathrm{~mm}

Thickness of each plate =12 \mathrm{~mm}

f_{y}=250 \mathrm{MPa}, f_{u}=410 \mathrm{MPa}

\begin{aligned} \gamma_{\mathrm{mw}} & =1.25 \\ \mathrm{P} & =275 \mathrm{kN} \text { (Factored) } \end{aligned}

Let us assume minimum length required of each weld =l

\Rightarrow \frac{\mathrm{f}_{\mathrm{u}}}{\sqrt{3} \gamma_{\mathrm{mw}}} \times 2 \mathrm{l} \times 0.7 \times size of weld =275 \times 10^{3}

\Rightarrow \quad \quad l=\frac{275 \times 10^{3} \times \sqrt{3} \times 1.25}{410 \times 2 \times 0.7 \times 10} =103.73 \mathrm{~mm}

The answer has to be round off to be nearest higher multiple of 5 and hence it shall be taken as 105 \mathrm{~mm}.

Hence, correct option is (B).

Thickness of each plate =12 \mathrm{~mm}

f_{y}=250 \mathrm{MPa}, f_{u}=410 \mathrm{MPa}

\begin{aligned} \gamma_{\mathrm{mw}} & =1.25 \\ \mathrm{P} & =275 \mathrm{kN} \text { (Factored) } \end{aligned}

Let us assume minimum length required of each weld =l

\Rightarrow \frac{\mathrm{f}_{\mathrm{u}}}{\sqrt{3} \gamma_{\mathrm{mw}}} \times 2 \mathrm{l} \times 0.7 \times size of weld =275 \times 10^{3}

\Rightarrow \quad \quad l=\frac{275 \times 10^{3} \times \sqrt{3} \times 1.25}{410 \times 2 \times 0.7 \times 10} =103.73 \mathrm{~mm}

The answer has to be round off to be nearest higher multiple of 5 and hence it shall be taken as 105 \mathrm{~mm}.

Hence, correct option is (B).

Question 3 |

Creep of concrete under compression is defined as the

increase in the magnitude of strain under
constant stress | |

increase in the magnitude of stress under
constant strain | |

decrease in the magnitude of strain under
constant stress | |

decrease in the magnitude of stress under
constant strain |

Question 3 Explanation:

Under sustained compressive loading, deformation in concrete increases with time even through the applied stress level is not changed. The time dependent component of strain is called creep.

Hence, correct answer is (A).

Hence, correct answer is (A).

Question 4 |

In the context of elastic theory of reinforced concrete, the modular ratio is
defined as the ratio of

Young's modulus of elasticity of reinforcement material to Young?s modulus of elasticity of concrete. | |

Youngs modulus of elasticity of concrete to Young?s modulus of elasticity of reinforcement material. | |

shear modulus of reinforcement material to the shear modulus of concrete. | |

Young's modulus of elasticity of reinforcement material to the shear modulus of concrete. |

Question 4 Explanation:

This is a question of working stress method i.e. elastic theory.

Modular ratio

=\frac{E_s}{E_c}=\frac{\text{Young's modulus of steel}}{\text{Young's modulus of concrete}}

Modular ratio

=\frac{E_s}{E_c}=\frac{\text{Young's modulus of steel}}{\text{Young's modulus of concrete}}

Question 5 |

The maximum applied load on a cylindrical concrete specimen of diameter 150 mm and
length 300 mm tested as per the split tensile strength test guidelines of IS 5816 : 1999
is 157 kN. The split tensile strength (in MPa, round off to one decimal place) of the
specimen is _______.

1.4 | |

2.2 | |

4.2 | |

6.4 |

Question 5 Explanation:

P = 157 kN

D = 150 mm

L = 300 mm

In split tensile strength test, split tensile

strength of concrete

\begin{aligned} f_{et}&=\frac{2P}{\pi DL}=\frac{2 \times 157000}{\pi \times 150 \times 300}\\ &=2.22 N/mm^2 \end{aligned}

There are 5 questions to complete.

Max.value of slump is used for what type of work?

thanks for the website

it helped me a lot