SSC JE Civil Engineering 2018 Set 2 Solved Paper

Question : 148 of 200

Marks: +1, -0

Calculate the design rate of super elevation on a 3 degree horizontal curve for a narrow gauge track, if design velocity on the curve is 70 km/h.

Solution:

R= 1720/ degree of curve = 1720/3 = 573.33m
For narrow gauge, G=0.762m
Super elevation (e) = GV

^{2}

/127R = 0.762 × 70

^{^2}

/ 127 × 573.33 = 0.051279

Go to Question:

Prev Question

Next Question