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

Question 6 |

Average free flow speed and the jam density observed on a roadstretch are 60 km/h and 120 vehicles/km, respectively. For a linear speed-density relationship, the maximum flow on the road stretch (in vehicles/h) is _________

900 | |

1800 | |

2700 | |

3600 |

Question 6 Explanation:

Maximum Speed / Free ?ow speed, V_m = 60 kmph

Jam density, K_m = 120 kmph

for linear speed - density (Green shield's model) relationship,

max. ?ow, is q_{max}=\frac{k_j}{2}\times \frac{V_f}{2}

q_{max}=\frac{120}{2}\times \frac{60}{2}=1800veh/hr

Jam density, K_m = 120 kmph

for linear speed - density (Green shield's model) relationship,

max. ?ow, is q_{max}=\frac{k_j}{2}\times \frac{V_f}{2}

q_{max}=\frac{120}{2}\times \frac{60}{2}=1800veh/hr

Question 7 |

Given the following data: design life n = 15 years, lane distribution factor D = 0.75, annual rate of growth of commercial vehicles r = 6%, vehicle damage factor F = 4 and initial traffic in the year of completion of construction = 3000 Commercial Vehicles Per Day (CVPD). As per IRC:37-2012, the design traffic in terms of cumulative number of standard axles (in million standard axles, up to two decimal places) is ______

25.12 | |

76.45 | |

100.78 | |

150.67 |

Question 7 Explanation:

\begin{aligned} N_{s}=& \frac{365 A\left[\left(1+\frac{r}{100}\right)^{n}-1\right] \times V D F \times L D F}{\frac{r}{100}} \\ =& \frac{365 \times 3000\left[1.06^{15}-1\right] \times 4 \times 0.75}{0.06} \times 10^{-6}\\ &76.45\text{msa} \end{aligned}

Question 8 |

The radii of relative stiffness of the rigid pavements P and Q are denoted by l_{P} and l_{Q}, respectively. The geometric and material properties of the concrete slab and underlying soil are given below:

The ratio (up to one decimal place) of l_{P} / l_{Q} is _______

The ratio (up to one decimal place) of l_{P} / l_{Q} is _______

8 | |

4 | |

6 | |

2 |

Question 8 Explanation:

Relative stiffness,

\begin{aligned} l &=\left[\frac{E h^{3}}{12 k\left(1-\mu^{2}\right)}\right]^{1 / 4} \\ \Rightarrow \quad \frac{l_{P}}{l_{Q}} &=\frac{\left(h^{3} / k\right)^{1 / 4}}{\left[\frac{(0.5 h)^{3}}{2 k}\right]^{1 / 4}}=2 \end{aligned}

\begin{aligned} l &=\left[\frac{E h^{3}}{12 k\left(1-\mu^{2}\right)}\right]^{1 / 4} \\ \Rightarrow \quad \frac{l_{P}}{l_{Q}} &=\frac{\left(h^{3} / k\right)^{1 / 4}}{\left[\frac{(0.5 h)^{3}}{2 k}\right]^{1 / 4}}=2 \end{aligned}

Question 9 |

In the context of the IRC 58-2011 guidelines for rigid pavement design, consider the following pair of statements.

I: Radius of relative stiffness is directly related to modulus of elasticity of concrete and inversely related to Poisson's ratio

II: Radius of relative stiffness is directly related to thickness of slab and modulus of subgrade reaction.

Which one of the following combinations is correct?

I: Radius of relative stiffness is directly related to modulus of elasticity of concrete and inversely related to Poisson's ratio

II: Radius of relative stiffness is directly related to thickness of slab and modulus of subgrade reaction.

Which one of the following combinations is correct?

I: True; II: True | |

I: False; II: False | |

I: True; II: False | |

I: False; II: True |

Question 9 Explanation:

Radius of relative stiffness,

l=\left[\frac{E h^{3}}{12 k\left(1-\mu^{2}\right)}\right]^{1 / 4}

As \mu increases then 1-\mu^{2} decreases

\therefore \quad Relative stiffness also increases.

\therefore Statement I is false.

Modulus of subgrade reaction,

k=\frac{P}{\Delta}

Statement II is false.

l=\left[\frac{E h^{3}}{12 k\left(1-\mu^{2}\right)}\right]^{1 / 4}

As \mu increases then 1-\mu^{2} decreases

\therefore \quad Relative stiffness also increases.

\therefore Statement I is false.

Modulus of subgrade reaction,

k=\frac{P}{\Delta}

Statement II is false.

Question 10 |

A traffic survey conducted on a road yields an average daily traffic count of 5000 vehicles. The axle load distribution on the same road is given in the following table:

The design period of the road is 15 years, the yearly traffic growth rate is 7.5% and the load safety factor(LSF)1.3. If the vehicle damage factor (VDF) is calculated from the above data, the design traffic (in million standard axle load, MSA) is ____________

The design period of the road is 15 years, the yearly traffic growth rate is 7.5% and the load safety factor(LSF)1.3. If the vehicle damage factor (VDF) is calculated from the above data, the design traffic (in million standard axle load, MSA) is ____________

212 | |

309 | |

424 | |

530 |

Question 10 Explanation:

Vehicle damage factor

\begin{aligned} &=0.1\left[\frac{18}{8.2}\right]^{4}+0.2\left[\frac{14}{8.2}\right]^{4}+0.35\left[\frac{10}{8.2}\right]^{4} \\ &\quad+0.15\left[\frac{8}{8.2}\right]^{4}+0.20\left[\frac{6}{8.2}\right]^{4} \\ N_{s}&=\frac{365 A\left((1+r)^{n}-1\right) D F}{r} \\ &=\frac{365 \times 5000\left((1.075)^{15}-1\right) \times 1.3 \times 4.988}{0.075} \\ N_{s}&=309.08 \mathrm{msa} \end{aligned}

\begin{aligned} &=0.1\left[\frac{18}{8.2}\right]^{4}+0.2\left[\frac{14}{8.2}\right]^{4}+0.35\left[\frac{10}{8.2}\right]^{4} \\ &\quad+0.15\left[\frac{8}{8.2}\right]^{4}+0.20\left[\frac{6}{8.2}\right]^{4} \\ N_{s}&=\frac{365 A\left((1+r)^{n}-1\right) D F}{r} \\ &=\frac{365 \times 5000\left((1.075)^{15}-1\right) \times 1.3 \times 4.988}{0.075} \\ N_{s}&=309.08 \mathrm{msa} \end{aligned}

There are 10 questions to complete.