Question 1 |

For an axle load of 15 tonne on a road, the Vehicle Damage Factor (round off to two
decimal places), in terms of the standard axle load of 8 tonne, is _________.

18.24 | |

24.12 | |

12.35 | |

16.25 |

Question 1 Explanation:

Axle load = 15 T

Standard axle load = 8 T

VDF=\left [ \frac{15}{8} \right ]^4=12.35

Standard axle load = 8 T

VDF=\left [ \frac{15}{8} \right ]^4=12.35

Question 2 |

A dowel bar is placed at a contraction joint. When contraction occurs, the concrete slab
cracks at predetermined location(s). Identify the arrangement, which shows the correct
placement of dowel bar and the place of occurrence of the contraction crack(s).

A | |

B | |

C | |

D |

Question 3 |

A road in a hilly terrain is to be laid at a gradient of 4.5%. A horizontal curve of radius
100 m is laid at a location on this road. Gradient needs to be eased due to combination
of curved horizontal and vertical profiles of the road. As per IRC, the compensated
gradient (in %, round off to one decimal place), is ______

2.2 | |

3.1 | |

4.8 | |

4 |

Question 3 Explanation:

Gradient = 4.5%, R = 100 m

Grade compensation

\begin{aligned} &=\left ( \frac{30+R}{R} \right )\ngtr \left ( \frac{75}{R} \right )\%\\ &=\frac{30+100}{100}\ngtr \frac{75}{100}\\ &=1.3\%\ngtr 0.75\; \\ & G.C>=0.75 \end{aligned}

Compansated Gradient = Gradient G.C = 4.5% - 0.75 = 3.75 \nless 4%

Hence C.G = 4%

Grade compensation

\begin{aligned} &=\left ( \frac{30+R}{R} \right )\ngtr \left ( \frac{75}{R} \right )\%\\ &=\frac{30+100}{100}\ngtr \frac{75}{100}\\ &=1.3\%\ngtr 0.75\; \\ & G.C>=0.75 \end{aligned}

Compansated Gradient = Gradient G.C = 4.5% - 0.75 = 3.75 \nless 4%

Hence C.G = 4%

Question 4 |

A flexible pavement has the following class of loads during a particular hour of the day.

i. 80 buses with 2-axles (each axle load of 40 kN);

ii. 160 trucks with 2-axles (front and rear axle loads of 40 kN and 80 kN, respectively)

The equivalent standard axle load repetitions for this vehicle combination as per IRC:37-2012 would be

i. 80 buses with 2-axles (each axle load of 40 kN);

ii. 160 trucks with 2-axles (front and rear axle loads of 40 kN and 80 kN, respectively)

The equivalent standard axle load repetitions for this vehicle combination as per IRC:37-2012 would be

180 | |

240 | |

250 | |

320 |

Question 4 Explanation:

(i) 80 buses with 2 axle with 40 kN each.

N_1=80 \times 2=160

L_1=40kN

160 \text{ truks} \left\{\begin{matrix} \text{Front axle}\rightarrow 40kN\\ \text{Rear axle}\rightarrow 80kN \end{matrix}\right.

N_1=160,\;L_1=40kN

N_2=160,\;L_2=80kN

\text{Total no. of repetitions}\left\{\begin{matrix} &\text{For}\; 40kN=160+160=320\\ &\text{For}\; 80kN=160 \end{matrix}\right.

N_1=320,\;L_1=40kN

N_2=160,\;L_2=80kN

As per 4th power law

\begin{aligned} N_s&=N_1\left ( \frac{L_1}{L_5} \right )^4 +N_2 \left ( \frac{L_2}{L_5} \right )^4 \\ &= 320\left ( \frac{40}{80} \right )^4+160\left ( \frac{80}{80} \right )^4\\ &= 20+160=180 \end{aligned}

N_1=80 \times 2=160

L_1=40kN

160 \text{ truks} \left\{\begin{matrix} \text{Front axle}\rightarrow 40kN\\ \text{Rear axle}\rightarrow 80kN \end{matrix}\right.

N_1=160,\;L_1=40kN

N_2=160,\;L_2=80kN

\text{Total no. of repetitions}\left\{\begin{matrix} &\text{For}\; 40kN=160+160=320\\ &\text{For}\; 80kN=160 \end{matrix}\right.

N_1=320,\;L_1=40kN

N_2=160,\;L_2=80kN

As per 4th power law

\begin{aligned} N_s&=N_1\left ( \frac{L_1}{L_5} \right )^4 +N_2 \left ( \frac{L_2}{L_5} \right )^4 \\ &= 320\left ( \frac{40}{80} \right )^4+160\left ( \frac{80}{80} \right )^4\\ &= 20+160=180 \end{aligned}

Question 5 |

A parabolic vertical curve is being designed to join a road of grade +5% with a road of grade -3%. The length of the vertical curve is 400m measured along the horizontal. The vertical point of curvature (VPC) is located on the road of grade +5%. The difference in height between VPC and vertical point of intersection (VPI) (in m, round off to the nearest integer)is _____

5 | |

10 | |

15 | |

20 |

Question 5 Explanation:

Given, N_1=+5\%,\; N_2=-3\%

Length of summit curve is, l = 400 m

Vertical distance between VPC and VPI is N_1 \times \frac{l}{2}=\frac{5\%}{100}\times 200=10m

Length of summit curve is, l = 400 m

Vertical distance between VPC and VPI is N_1 \times \frac{l}{2}=\frac{5\%}{100}\times 200=10m

There are 5 questions to complete.