Question 1 |
The lane configuration with lane volumes in vehicles per hour of a four-arm
signalized intersection is shown in the figure. There are only two phases: the first
phase is for the East-West and the West-East through movements, and the second
phase is for the North-South and the South-North through movements. There are
no turning movements. Assume that the saturation flow is 1800 vehicles per hour
per lane for each lane and the total lost time for the first and the second phases
together is 9 seconds.
The optimum cycle length (in seconds), as per the Webster's method, is ____________. (round off to the nearest integer)
The optimum cycle length (in seconds), as per the Webster's method, is ____________. (round off to the nearest integer)
12 | |
45 | |
37 | |
65 |
Question 1 Explanation:
From figure
\begin{aligned} y_{N-S}&=\left [ \frac{360}{1800},\frac{396}{1800} \right ]_{max}\\ &=0.22\\ y_{E-W}&=\left [ \frac{504+504}{2 \times 18},\frac{440}{1800} ,\frac{460}{1800}\right ]_{max}\\ &=0.28\\ &\text{Optimum cycle length}\\ &=\frac{1.5L+5}{1-y}\\ &=\frac{1.5 \times 9+5}{1-(y_{N-S}+y_{E-W})}\\ &=\frac{1.5 \times 9+5}{1-(0.22+0.28)}\\ C_o&=37secs. \end{aligned}
\begin{aligned} y_{N-S}&=\left [ \frac{360}{1800},\frac{396}{1800} \right ]_{max}\\ &=0.22\\ y_{E-W}&=\left [ \frac{504+504}{2 \times 18},\frac{440}{1800} ,\frac{460}{1800}\right ]_{max}\\ &=0.28\\ &\text{Optimum cycle length}\\ &=\frac{1.5L+5}{1-y}\\ &=\frac{1.5 \times 9+5}{1-(y_{N-S}+y_{E-W})}\\ &=\frac{1.5 \times 9+5}{1-(0.22+0.28)}\\ C_o&=37secs. \end{aligned}
Question 2 |
Assuming that traffic on a highway obeys the Greenshields model, the speed of
a shockwave between two traffic streams (P) and (Q) as shown in the schematic
is _______ kmph. (in integer)
5 | |
10 | |
15 | |
25 |
Question 2 Explanation:
\text{Speed of shock wave}=\frac{\text{Change in flow}}{\text{Change in density}}
\text{Flow}=\text{Speed} \times \text{Density}
\begin{aligned} Density&=\frac{Flow}{speed}\\ &=\frac{q_Q-q_P}{\frac{q_Q}{V_Q}-\frac{q_P}{V_P}}\\ &=\frac{1800-1200}{\frac{1800}{30}-\frac{1200}{60}}\\ &=\frac{600}{60-20}\\ &=15Kmph \end{aligned}
\text{Flow}=\text{Speed} \times \text{Density}
\begin{aligned} Density&=\frac{Flow}{speed}\\ &=\frac{q_Q-q_P}{\frac{q_Q}{V_Q}-\frac{q_P}{V_P}}\\ &=\frac{1800-1200}{\frac{1800}{30}-\frac{1200}{60}}\\ &=\frac{600}{60-20}\\ &=15Kmph \end{aligned}
Question 3 |
A parabolic vertical crest curve connects two road segments with grades +1.0% and -2.0%. If a 200 m stopping sight distance is needed for a driver at a height of 1.2 m to avoid an obstacle of height 0.15 m, then the minimum curve length
should be ______ m. (round off to the nearest integer)
241 | |
365 | |
115 | |
273 |
Question 3 Explanation:
Given that, n_1=+1% and n_2=-2%
n=n_1-n_2=3%
SSD = 200 m
and h_1= 1.2 m and h_2= 0.15 m
as given n_1 up gradient, and n_2 - down gradient.
So curve is summit curve.
Assume L > SSD
\begin{aligned} L &=\frac{NS^2}{2(\sqrt{n_1}+\sqrt{n_2})^2} \\ &= \frac{3}{100} \times \frac{(200)^2}{2 \times (\sqrt{1.2}+\sqrt{0.15})^2}\\ &=272.91>200 \\ L&=272.91m \end{aligned}
n=n_1-n_2=3%
SSD = 200 m
and h_1= 1.2 m and h_2= 0.15 m
as given n_1 up gradient, and n_2 - down gradient.
So curve is summit curve.
Assume L > SSD
\begin{aligned} L &=\frac{NS^2}{2(\sqrt{n_1}+\sqrt{n_2})^2} \\ &= \frac{3}{100} \times \frac{(200)^2}{2 \times (\sqrt{1.2}+\sqrt{0.15})^2}\\ &=272.91>200 \\ L&=272.91m \end{aligned}
Question 4 |
A single-lane highway has a traffic density of 40 vehicles/km. If the time-mean
speed and space-mean speed are 40 kmph and 30 kmph, respectively, the
average headway (in seconds) between the vehicles is
3 | |
2.25 | |
8.33 \times 10^{-4} | |
6.25 \times 10^{-4} |
Question 4 Explanation:
Given that,
Traffic density = 40 Veh/Km.
Time mean speed = 40 Kmph
Space mean speed = 30 kmph
Time Headway (sec) = ?
We know that traffic volume
= Density x Space mean speed = 40x30 =1200 Veh hr.
and we also know that
q=\frac{3600}{H_t}
where H_t- average headway (sc)
1200=\frac{3600}{H_t}
H_t=\frac{3600}{1200}=3 secs.
Traffic density = 40 Veh/Km.
Time mean speed = 40 Kmph
Space mean speed = 30 kmph
Time Headway (sec) = ?
We know that traffic volume
= Density x Space mean speed = 40x30 =1200 Veh hr.
and we also know that
q=\frac{3600}{H_t}
where H_t- average headway (sc)
1200=\frac{3600}{H_t}
H_t=\frac{3600}{1200}=3 secs.
Question 5 |
The base length of the runway at the mean sea level (MSL) is 1500 m. If the
runway is located at an altitude of 300 m above the MSL, the actual length
(in m) of the runway to be provided is ____________. (round off to the nearest
integer)
1245 | |
2354 | |
1605 | |
1248 |
Question 5 Explanation:
Correction for elevation = It should increase at a
rate of 7% per 300 m rise in elevation from MSL.
Given that
Basic runway length as MSL =1500m
Elevation =300m
Correction =\frac{7}{100}\times \frac{300}{300}\times 1500=105m
The actual length of runway= 1500+150=1605m
Given that
Basic runway length as MSL =1500m
Elevation =300m
Correction =\frac{7}{100}\times \frac{300}{300}\times 1500=105m
The actual length of runway= 1500+150=1605m
Question 6 |
For a traffic stream, v is the space mean speed, k is the density, q is the flow,
v_f is the free flow speed, and k_j is the jam density. Assume that the speed
decreases linearly with density.
Which of the following relation(s) is/are correct?
Which of the following relation(s) is/are correct?
q=k_jk-\left ( \frac{k_j}{v_f} \right )k^2 | |
q=v_fk-\left ( \frac{v_f}{k_j} \right )k^2 | |
q=v_fv-\left ( \frac{v_f}{k_j} \right )v^2 | |
q=k_jv-\left ( \frac{k_j}{v_f} \right )v^2 |
Question 6 Explanation:
As per Greenshield's
\begin{aligned} V&=V_f\left ( 1-\frac{K}{K_j} \right )\\ q&=K \times V \\ &= K \left [ 1-\frac{K}{K_j} \right ]V_f\\ q&=V_fK-\left ( \frac{V_f}{K_j} \right )K^2\\ V&=V_f\left [ 1-\frac{K}{K_j} \right ] \end{aligned}
From above equation we can say that
\begin{aligned} K&=K_j\left ( 1-\frac{V}{V_f} \right )\\ q&=K \times V \\ q&= K_j \left [ 1-\frac{V}{V_f} \right ] \times V\\ q&=K_jV-\left ( \frac{K_j}{V_f} \right )V^2 \end{aligned}
\begin{aligned} V&=V_f\left ( 1-\frac{K}{K_j} \right )\\ q&=K \times V \\ &= K \left [ 1-\frac{K}{K_j} \right ]V_f\\ q&=V_fK-\left ( \frac{V_f}{K_j} \right )K^2\\ V&=V_f\left [ 1-\frac{K}{K_j} \right ] \end{aligned}
From above equation we can say that
\begin{aligned} K&=K_j\left ( 1-\frac{V}{V_f} \right )\\ q&=K \times V \\ q&= K_j \left [ 1-\frac{V}{V_f} \right ] \times V\\ q&=K_jV-\left ( \frac{K_j}{V_f} \right )V^2 \end{aligned}
Question 7 |
For the dual-wheel carrying assembly shown in the figure, P is the load on each
wheel, a is the radius of the contact area of the wheel, s is the spacing between
the wheels, and d is the clear distance between the wheels. Assuming that the
ground is an elastic, homogeneous, and isotropic half space, the ratio of
Equivalent Single Wheel Load (ESWL) at depth z=d/2 to the ESWL at depth
z=2s is ___________. (round off to one decimal place)
(Consider the influence angle to be 45^{\circ} for the linear dispersion of stress with
depth)
0.2 | |
0.5 | |
0.8 | |
0.1 |
Question 7 Explanation:
\begin{aligned}
\text{ESWL @ depth }\frac{d}{2}&=P \\
\text{ESWL @ depth }2S&=2P \\
\text{Ratio }\frac{P}{2P}&=0.5 \\
\end{aligned}
Question 8 |
The vehicle count obtained in every 10 minute interval of a traffic volume survey done in peak one hour is given below.
\begin{array}{|c|c|}\hline \text{Time Interval}& \text{Vehicle Count} \\ \text{(in minutes)}& \\ \hline 0-10& 10\\ \hline 10-20 &11\\ \hline 20-30&12\\ \hline 30-40&15\\ \hline 40-50 & 13\\ \hline 50-60 &11\\ \hline \end{array}
The peak hour factor (PHF) for 10 minute sub-interval is __________. (round off to one decimal place)
\begin{array}{|c|c|}\hline \text{Time Interval}& \text{Vehicle Count} \\ \text{(in minutes)}& \\ \hline 0-10& 10\\ \hline 10-20 &11\\ \hline 20-30&12\\ \hline 30-40&15\\ \hline 40-50 & 13\\ \hline 50-60 &11\\ \hline \end{array}
The peak hour factor (PHF) for 10 minute sub-interval is __________. (round off to one decimal place)
0.2 | |
0.4 | |
0.8 | |
0.1 |
Question 8 Explanation:
\begin{aligned}
PHF&=\frac{\text{Peak flow during 1hour}}{6 \times \text{Peak flow during 10 minutes}}\\
&=\frac{10+11+12+15+13+11}{6 \times 15}\\
&=\frac{72}{6 \times 15}\\
&=0.8
\end{aligned}
Question 9 |
At a traffic intersection, cars and buses arrive randomly according to independent
Poisson processes at an average rate of 4 vehicles per hour and 2 vehicles per
hour, respectively. The probability of observing at least 2 vehicles in 30 minutes
is ______. (round off to two decimal places)
0.52 | |
0.64 | |
0.80 | |
0.44 |
Question 9 Explanation:
\begin{aligned}
\lambda _C&=4 \; vehical/hr=2 \; vehical /30 \;min\\
\lambda _C&=2 \; vehical/hr=1 \; vehical /30 \;min\\
\lambda _{Vehical}&=3 \; vehical/30 \;min\\
P(x \lt 2)&=1-P(x\leq 1)\\
&=1-[P(x=0)+P(x=1)]\\
&=1-\left [ \frac{e^{-\lambda }\lambda ^0}{0!} +\frac{e^{-\lambda }\lambda ^1}{1!}\right ]\\
&=1-e^{-3}(1+3)\\
&=1-\frac{4}{e^3}\\
&=0.8
\end{aligned}
Question 10 |
A two-phase signalized intersection is designed with a cycle time of 100 s. The
amber and red times for each phase are 4 s and 50 s, respectively. If the total
lost time per phase due to start-up and clearance is 2 s, the effective green time
of each phase is ______s. (in integer)
96 | |
56 | |
82 | |
48 |
Question 10 Explanation:
Two phase signalized intersections,
Cycle time = 100 secs.
Amber time = 4 secs. [each phase]
Red time = 50 secs. [each phase]
Total lost time = 2 sec. [each phase]
Effective green time = ? [each phase]
Total Effective green = Cycle time - Lost time = 100-2x2 = 96 sec.
As there is symmetry in phases so effective green time per phase= 96/2=48 sec.
Cycle time = 100 secs.
Amber time = 4 secs. [each phase]
Red time = 50 secs. [each phase]
Total lost time = 2 sec. [each phase]
Effective green time = ? [each phase]
Total Effective green = Cycle time - Lost time = 100-2x2 = 96 sec.
As there is symmetry in phases so effective green time per phase= 96/2=48 sec.
There are 10 questions to complete.