Courses

Railway & Airport Engineering Civil Engineering (CE) Notes | EduRev

Topic wise GATE Past Year Papers for Civil Engineering

Created by: Gate Gurus

Civil Engineering (CE) : Railway & Airport Engineering Civil Engineering (CE) Notes | EduRev

The document Railway & Airport Engineering Civil Engineering (CE) Notes | EduRev is a part of the Civil Engineering (CE) Course Topic wise GATE Past Year Papers for Civil Engineering.
All you need of Civil Engineering (CE) at this link: Civil Engineering (CE)

Question 1. A broad gauge railway line passes through a horizontal curved section (radius = 875 m) of length 200 m. The allowable speed on this portion is 100 km/h. For calculating the cant, consider the gauge as centre-to-centre distance between the rail heads, equal to 1750 mm. The maximum permissible cant (in mm, round off to 1 decimal place) with respect to the centre-to-centre distance between the rail heads is_______      [2019 : 2 Marks, Set-II]
Solution:

Question 2. For a broad gauge railway track on a horizontal curve of radius R (in m), the equilibrium cant e required for a train moving at a speed of V(in km per hour) is    [2017 : 1 Mark, Set-II]
(a)
(b)
(c)
(d)

Solution:

Question 3. A runway is being constructed in a new airport as per the International Civil Aviation Organization (ICAO) recommendations. The elevation and the airport reference temperature of this airport are 535 m above the mean sea level and 22.65Â°C, respectively. Consider the effective gradient of runway as 1%. The length of runway required for a design-aircraft under the standard conditions is 2000 m. Within the framework of applying sequential corrections as per the ICAO recommendations, the length of runway corrected for the temperature is     [2017 : 1 Mark, Set-I]
(a) 2223 m
(b) 2250 m
(c) 2500 m
(d) 2750 m
Solution: Elevation = 535 m
Airport Reference Temperature = 22.65Â° C
Runway length = 2000 m under standard conditions.
Runway length correction for TÂ°C.
=> Correction for elevation

Corrected runway length = 2249.67 m
=> Corrected standard TÂ° at 535 m elevation
= 15Â°C - (0.0065 x 535) = 11.52Â°C
Correction for TÂ°

Corrected Runway length
= 2249.67 + 250.39

Combined elevation and temperature correction
= 500 m < (0.35 x 2000)

Offer running on EduRev: Apply code STAYHOME200 to get INR 200 off on our premium plan EduRev Infinity!

69 docs

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

;