# Pavement Design

 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 _________.
 A 18.24 B 24.12 C 12.35 D 16.25
GATE CE 2020 SET-2   Transportation Engineering
Question 1 Explanation:
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 A B B C C D D
GATE CE 2020 SET-1   Transportation Engineering
 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 ______
 A 2.2 B 3.1 C 4.8 D 4
GATE CE 2020 SET-1   Transportation Engineering
Question 3 Explanation:
Gradient = 4.5%, R = 100 m
\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
 A 180 B 240 C 250 D 320
GATE CE 2019 SET-2   Transportation Engineering
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}
 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 _____
 A 5 B 10 C 15 D 20
GATE CE 2019 SET-1   Transportation Engineering
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 $N_1 \times \frac{l}{2}=\frac{5\%}{100}\times 200=10m$
 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 _________
 A 900 B 1800 C 2700 D 3600
GATE CE 2019 SET-1   Transportation Engineering
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 $q_{max}=\frac{k_j}{2}\times \frac{V_f}{2}$
$q_{max}=\frac{120}{2}\times \frac{60}{2}=1800veh/hr$
 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 ______
 A 25.12 B 76.45 C 100.78 D 150.67
GATE CE 2018 SET-1   Transportation Engineering
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 _______
 A 8 B 4 C 6 D 2
GATE CE 2017 SET-2   Transportation Engineering
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}
 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?
 A I: True; II: True B I: False; II: False C I: True; II: False D I: False; II: True
GATE CE 2016 SET-2   Transportation Engineering
Question 9 Explanation:
$l=\left[\frac{E h^{3}}{12 k\left(1-\mu^{2}\right)}\right]^{1 / 4}$
As $\mu$ increases then $1-\mu^{2}$ decreases
$\therefore \quad$ Relative stiffness also increases.
$\therefore$ Statement I is false.
$k=\frac{P}{\Delta}$
\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}