7/30/2019 Railway Engineering Model Questions.pdf

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Masters of Science in transportation Engineering & Management

Railway Engineering Model Questions

1. Comparison of railway & road transport and advantages of railway transport.2. Railway classification based on gauge.3. What are the components of rolling stock of railway?4. What are the component parts of railway track? Neat sketch of the railway track cross-section.5. Types of gradients and what is the grade compensation on curve? Numerical example on grade

compensation.6. What are the requirements of a good track structure?7. Calculate the material requirements for one Km of railway track: Rails, number of sleepers, fish plates, fish

bolt, and bearing plate, stone ballast etc.8. What are the types of stations and their functions?9. What are the general requirements on terminal stations?10. What are the differences between platforms and yards?11. Determination of maximum permissible speed on curves12. Determination of transition curve length and length of vertical curve13. Determine superelevation (numerical)14. Determine maximum permissible speed (numerical), transition curve length15. Components of turn-out make a neat sketch of left hand turn-out.

Numerical:1. Materials Requirement per Km of railway track

Quantities of different materials required to lay one Km track can be calculate as follows:a) No. of rails per Km length = (1000x2)/rail length =

For BG Track 60 kg rail and 13 m length of rail recommended:No. of rails per km length = (2x100)/13=154

b) Weight of rail per km length = 154x13x60 = 120120 kg.c) No. of sleepers per km length = (No. of rails per km)/2 x (sleeper density)On BG track sleeper density = 13 + 7 = 20

No. of sleepers per km track = (154/2)x20 = 1540nosd) No of fish plates per one km of track length = no of rails per km x2 = 154x2 = 308 nose) No. of fish bolts = 4x no of rails per km = 4x154 = 616 nos.f) No of bearing plates = 2 x No of sleepers per km length = 2x 1540 = 3080 nos.g) No of labours required to lay one km of track (with 8 hour shift) = total tonnage + 20%

2. The ruling gradient on a BG track section has been fixed as 1 in 200. What should be the compensated gradientwhen a 4 degree horizontal curve is to be provided on this ruling gradient?

3. Calculate the maximum permissible on a curve of high speed BG route with the following data:a. Degree of curve = 1degreeb.

Superelevation = 85 mmc. Length of transition curve = 125m

d. Sanctioned speed of the section = 170 kmph4. On a BG route involving high speed, A 100 m transition curve has been provided and a superelevation of 80mm

has been managed. The degree of curve is 1 and the maximum sanctioned speed for the curved section is 170kmph. Determine maximum permissible speed on the curve.

(Hint: assume cant deficiency as 75 mm)

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5. Find out the superelevation to be provided and the maximum permissible speed for 2 degree BG transition curveon a high speed route having a maximum sanctioned speed of the section as 100 kmph. For calculating theequilibrium superelevation the speed given as 75 kmph and the booked speed for goods traffic is 50 kmph.

6. Calculate the superelevation and maximum permissible speed for a 30 curve on a high speed BG track with thefollowing data:

a. maximum sanctioned speed = 130 kmphb. equilibrium speed = 85 kmphc. booked speed for goods train = 50 kmph

7. Find out the length of transition curve for a four degree BG circular curved track having a cant of 15 cm. themaximum permissible speed on the curve is 90 kmph. Find out the shift and off-set at every 15 m interval of thecurve. Draw the transition curve also. Assume maximum permissible cant deficiency is 75 mm.

8. A rising gradient of 1 in 150 m meets a falling gradient of 1 in 250 on a group A route. The intersection pointhas a chainage of 1000m and its RL is 100 m Calculate following:

a. Length of the vertical curveb. RL of vertical length and chainage of various points

9.