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 |

With regard to the shear design of RCC beams, which of the following statements is/are TRUE?

Excessive shear reinforcement can lead to compression failure in concrete | |

Beams without shear reinforcement, even if adequately designed for flexure, can have brittle failure | |

The main (longitudinal) reinforcement plays no role in the shear resistance of beam | |

As per IS456:2000, the nominal shear stress in the beams of varying depth depends on both the design shear force as well as the design bending moment |

Question 3 Explanation:

Option A is true, because when the area of shear reinforcement is large i.e. in case of excessive shear reinforcement, concrete becomes stronger is diagonal tension on failure compared to diagonal compression failure and compression failure may occur before the shear reinforcement has yielded.

Option 'B' is true because for beams without shear reinforcement, once the flexural crack crosses the longitudinal reinforcement, the propagation of crack will be sudden and there can be brittle failure.

Option 'C' is not true, as main reinforcement increases the shear resitance by providing dowel action, limiting crack width and by increasing the depth of concrete. Design shear strength of concrete is a function of grade of concree and percentage tensile reinforcement.

Option 'D' is true, as for beams with varying depth, nominal shear stress,

\tau_{v}=\frac{V_{u} \pm \frac{M_{u} \tan \beta}{d}}{b d}( clause 40.1.1, IS456: 2000)

it is dependent on both the design shear force and design bending moment.

Hence, options (A), (B) and (D) are true.

Option 'B' is true because for beams without shear reinforcement, once the flexural crack crosses the longitudinal reinforcement, the propagation of crack will be sudden and there can be brittle failure.

Option 'C' is not true, as main reinforcement increases the shear resitance by providing dowel action, limiting crack width and by increasing the depth of concrete. Design shear strength of concrete is a function of grade of concree and percentage tensile reinforcement.

Option 'D' is true, as for beams with varying depth, nominal shear stress,

\tau_{v}=\frac{V_{u} \pm \frac{M_{u} \tan \beta}{d}}{b d}( clause 40.1.1, IS456: 2000)

it is dependent on both the design shear force and design bending moment.

Hence, options (A), (B) and (D) are true.

Question 4 |

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 4 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 5 |

Consider a doubly reinforced RCC beam with the option of using either Fe250 plain bars or Fe500 deformed bars in the compression zone. The modulus of elasticity of steel is 2 \times 10^{5} \mathrm{~N} / \mathrm{mm}^{2}. As per IS456-2000, in which type(s) of the bars, the stress in the compression steel \left(f_{s c}\right) can reach the design strength \left(0.87 f_{y}\right) at the limit state of collapse?

Fe250 plain bars only | |

Fe500 deformed bars only | |

Both Fe250 plain bars and Fe500 deformed
bars
| |

Neither Fe250 plain bars nor Fe500 deformed
bars |

Question 5 Explanation:

In a doubly reinforced beam, at the limit state of collapse. The strain in the extreme corression fibre will be 0.0035. As the compression reinforcement will be below this, the strain in if will be always less than 0.0035.

For Fe250, the yield strain

=\frac{0.87 \times 250}{2 \times 10^{5}}=0.0010875

For Fe500, the yield strain

\begin{aligned} & =0.002+\frac{0.87 \mathrm{f}_{y}}{E_{s}} \\ & =0.002+\frac{0.87 \times 500}{2 \times 10^{5}} \\ & =0.004175>0.0035 \end{aligned}

So, it is possible that \mathrm{Fe} 250 can reach upto its yield strain and hence can reach the design strength 0.87 \mathrm{f}_{\mathrm{y}}.

For Fe250, the yield strain

=\frac{0.87 \times 250}{2 \times 10^{5}}=0.0010875

For Fe500, the yield strain

\begin{aligned} & =0.002+\frac{0.87 \mathrm{f}_{y}}{E_{s}} \\ & =0.002+\frac{0.87 \times 500}{2 \times 10^{5}} \\ & =0.004175>0.0035 \end{aligned}

So, it is possible that \mathrm{Fe} 250 can reach upto its yield strain and hence can reach the design strength 0.87 \mathrm{f}_{\mathrm{y}}.

There are 5 questions to complete.

The answer of q.9 shall be 122 kN-m as it is under reinforced section the evaluation of moment of the resistance will be based on tension not the compression.