Geometric design of the track

As per vtu prescribed syllabus. Transposition Engineering-II
unit 3

1. GEOMETRIC DESIGN OF THE TRACK

2. Necessity of Geometric Design: Smooth & safe running of trains. Maximum speed. Carrying heavy axle loads. Avoid accidents & derailments. Less maintenance efforts. Good aesthetic value. There , if all the above elements are properly designed, the possibility of derailments due to defects in the track can be avoided.

3. Gradients The amount of slope in longitudinal direction of railway track is called gradient or grade. Gradients are provided to negotiate the rise or fall in the level of the railway track. Rising gradient rises the track in the direction of movement, whereas, falling gradient cause the track to go down in the direction of movement. A gradient is represented by the distance traveled for a rise or fall of one unit. It is written as; 1 in ‘X’ or 1 in ‘n’ or as percent

4. Gradients are provided on the track due to the following reasons. To provide a uniform rate of rise or fall as far as possible. To reach the various stations located at different elevation. To reduce the cost of earth work.

5. Gradients – Types Ruling gradient Momentum gradient Pusher or Helper gradient Gradients in Station Yards

6. A) Ruling gradient: This is the design gradient basically, because it is determined on the basis of the performance of the locomotive and at the same time, it tries to look at the total amount of load which that locomotive can take up along with it, while negotiating any gradient without any loss or major loss in the speed of the movement.

7. Ruling gradient; It is a maximum gradient (steepest gradient), which may be permitted on the section of the track. It is determined by maximum load that a locomotive can haul with maximum permissible speed. Extra pull required by locomotive on gradient with ‘ ’ inclination. P = W Sin = W * gradient For Ex:- Weight of train (W)=500 tonnes Gradient = 1 in 100 Extra power (P) = 5 tonnes

8. Ruling gradient with one locomotive. In Plane area = 1 in 150 to 1 in 250 In hilly area = 1 in 100 to 1in 150 Once a ruling gradient is specified for a section, then all other gradients provided in that section should be flatter than the ruling gradients (after making due to compensation for curvature)

9. Gradients – Types …. b ) Momentum gradient: Momentum g radient steeper than ruling gradient that is overcome by momentum gathered while having a run in plane or on falling gradient in valleys. Use additional kinetic energy received during run on a section. No obstruction like signals are provided on section with these gradients.(means the train should not be stopped at that territory)

10. For example, in valleys, a falling gradient is usually followed by rising gradient acquires sufficient momentum. This momentum gives additional kinetic energy to the moving train which would enable the train to overcome a steeper rising gradient than the ruling gradient for a certain length of the track. This rising gradient is called momentum gradient and this gradient is steeper than ruling gradient.

11. C) Pusher or Helper gradient: Gradient steeper than ruling gradient requiring extra locomotive. It reduces the length of a railway section. It also reduces the overall cost. Examples : In Darjeeling Railways 1 in 37 Pusher gradient is used on Western Ghats, B.G tracks & N.G tracks 1 in 25 is provided.

12. d) Gradient at Station Yards: As per as possible the track along the stations & yards should be level or gradients should be sufficiently low. To prevent standing vehicle from rolling & moving away from the yard due to combined effect of gravity & strong winds. To reduce additional resistive forces required to start a locomotive to the extent possible. Minimum gradient from drainage consideration. On Indian railways, maximum gradient permitted is 1 in 400 in station yards & minimum gradient permitted is 1 in 1000

13. Grade Compensation ( On Curves):- If a curve is provided on a track with ruling gradient, the resistance of the track will be increased on this curve. In order to avoid resistance beyond the allowable limits, the gradients are reduced on curves & this reduction in gradient is known as grade compensation for curves. In India, Compensation for curvature is given by. BG track :  0.04% per degree of curve MG track : 0.03 % per degree of curve NG track :  0.02 % per degree of curve

14. Grade Compensation ( On Curves); . Example :If the ruling gradient is 1 in 250 on a particular section of B.G & at the same time a curve of 4 degree is situated on this ruling gradient, what should be the allowable ruling gradient? Solution : As per Indian railway recommendation, the grade compensation for of B.G track is 0.04% per degree of curve. Therefore, Grade compensation for 4 degree curve = 0.04 * 4 = 0.16% Ruling gradients is 1 in 250 = 1/250 *100 = 0.4% Therefore, Required ruling gradient or Actual gradient = Ruling gradient – grade compensation = 0.4- 0.16 = 0.24% or 1 in 417

15. Radius & Degree of a curve:- The main curved portion of a railway track is kept circular i.e , the radius at every point of the curve is same. The radius of a railway curve is sometimes represented by the degree of the curve. Degree of a railway curve :- The angle subtended at the center of the curve by an arc of 30.0m length is defined as a defined as a degree of the curve. Relationship between the radius & degree of a curve . Let, R=radius of the curve in meters, D = degree of the curve Now, total circumference 2 π R makes 360 o at the center. Therefore, for 30.0 m arc makes an angle. D/30 = 360/2 π R D = 360*30/2 π R D = 1718.87/R D = 1720/R

16. Safe speed on curves.. . Safe speed for all practical purposes means a speed which is safe from the danger of overturning & derailment with a certain margin of safety. This speed, to negotiate curves safely, depends upon the following factors. The gauge of track. The radius of the curve. Amount of super elevation provided. In India, using safe speed V in kmph, formula. B.G & M.G : V = 4.35 R- 67 N.G: V = 3.65 R- 6

17. Superelevation or Cant …. When a train moves round a curve, it is subjected to a centrifugal force acting horizontally at the centre of gravity of each vehicle radially away from the centre of the curve. This increases the weight on the outer rails. To counteract the effect of centrifugal forces, the level of the outer rail is raised above the inner rail by a certain amount to introduce the centripetal force. This raised elevation of outer rail above the inner rail at a horizontal curve is called “Cant” .

18. Superelevation or Cant …. It is the difference in elevation (or height) between the outer rail and inner rail at a horizontal curve is called “Cant” . Inner rail is taken as a reference rail & is maintained at its original level. Inner rail is also known as ‘gradient rail’. It is denoted by “e”.

19. Superelevation or Cant …. Objectives of Superelevation. To neutralize the effect of centrifugal force. Equal distribution of wheel loads. Providing smooth track, improving passenger comfort. To reduce wear & tear of the rails & rolling stock.

20. Equilibrium Cant …. The cant or super elevation as given by equation e=(GV 2 /1.27R) cm, the load carried by both the wheels will be the same, the springs will be equally compressed & the passengers will not tend to lean in either direction, such cant is known as the “Equilibrium cant”.

21. Equilibrium Cant …. The cant is provided on the basis of average speed of the trains. The majority of Indian Railways provide super elevation for equilibrium speed or average speed under condition of level track. Average speed or Weighted average speed = n 1 V 1 + n 2 V 2 + n 3 V 3 / n 1 + n 2 + n 3 Where, n 1 , n 2, n 3 = Number of trains V 1 , V 2, V 3 = Speed of trains in kmph

22. Equilibrium Cant …. The cant is provided on the basis of average speed of the trains. The majority of Indian Railways provide super elevation for equilibrium speed or average speed under condition of level track. Average speed or Weighted average speed = n 1 V 1 + n 2 V 2 + n 3 V 3 / n 1 + n 2 + n 3 Where, n 1 , n 2, n 3 = Number of trains V 1 , V 2, V 3 = Speed of trains in kmph

23. Cant Deficiency (C d ) …. The equilibrium cant is provided on the basis of equilibrium speed (or Average speed) of different trains. But this equilibrium cant or super elevation falls short of that required for the high speed trains. This shortage of cant is called “ Cant Deficiency” . In other words, it is the difference between the equilibrium cant necessary for the maximum permissible speed on a curve and the actual cant provided. Higher cant deficiency causes more unbalanced centrifugal force and discomfort to the passengers. Maximum value of cant deficiency prescribed for Indian Railways . B.G = 7.6cm, M.G = 5.1cm, N.G = 3.8cm

24. Negative Superelevation…. When a branch line diverges from a main line on a curve of contrary flexure, the super elevation necessary for the average speed of trains running over the main line, cannot be provided. The speed of the trains over the diverging track and main line track has to be reduced considerably. The reason for the reduction of speed is that, on the branch line track, the inner rail remains at higher level than the outer rail.

25. Negative Superelevation…. D Outer rail Inner rail Inner rail

26. Negative Superelevation…. Ref fig : AD which is the outer rail of the main line curve must be higher than inner rail BC or in other words, the point A should be higher than point B . For the branch line, however, BE should be higher than AF or the point B should be higher than point A . These two contradictory conditions cannot be meet at the same time within one layout . So, outer rail BE on branch line is kept lower than the inner rail AF. In such case branch line curve has a negative super elevation & therefore speeds on both tracks must be restricted, particularly on branch line.

27. Negative Superelevation…. Calculation of restricted speed on the main line & branch line . S uper elevation for branch line can be calculated Calculate equilibrium super elevation for branch line (e b ). Find super elevation, X = e b – C d for branch line. This is to be provided on main line. To calculate the maximum permissible speed for main line, Cant should be, e m = X + C d S uper elevation for main line can be calculated Calculate equilibrium super elevation for main line ( e m ). Find super elevation, X = e m – C d for main line. Provided cant on branch line as ( -X ). To calculate the maximum permissible speed for branch line, Cant should be, e b = (-X ) + C d

28. Transitional Curves…. Transition curve is defined as a curve of parabolic nature which is introduced between a straight and a circular curve or between two branches of a compound curve. It is necessary to provide an easy change from a tangent to the radius selected for a particular curve . It is essential that the curvature and superelevation in the outer rail and the curvature in the inner rail are attained gradually, by the use of easement curve or transition curve.

29. Transitional Curves…. Objects : Primary objects : To decrease the radius of curve gradually from infinite at the straight to that of circular curve of selected radius. To attain gradual rise for the desired superelevation. This is applicable for outer rail. Secondary objects: The gradual increase or decrease of the centrifugal force on the vehicle by use of this curve provides smooth running of vehicle & comfort to the passengers. No sudden application, so the chances of derailment are gently reduced.

30. Transitional Curves…. Requirements of a Transition Curve ; It should be perfectly tangential to the straight. The length of the transition curve should be such that curvature may increase at the same rate as the superelevation . This curve should join the circular arc tangentially i.e., curvature of transition curve should conform with that of circular curve.

31. Superelevation for B.G, M.G & N.G…. Relationship of superelevation (e), with gauge(G), speed (V) and radius of the curve ( R ). W= Weight of moving vehicle in kg. v = Speed of vehicle in m/sec V = Speed of vehicle in km.ph R = Radius of curve in meters G = Gauge of track in meters g = Acceleration due to gravity in m/sec 2 α = Angle of inclination S = Length of inclined surface in meters Centrifugal force is given by the following expression. F = Wv 2 / gR -------------(1) Now resolving the forces along the inclined surface we get F cos α = W sin α -------- (2)

32. Superelevation for B.G, M.G & N.G…. F = Wv 2 / gR , cos α = G/S & sin α = e/S Therefore equation (2) becomes Wv 2 / gR * G/S = W*e/S Therefore, e = v 2 / gR * G meters Where, v in m/sec. = G ( 0,278V) 2 /9.81R = GV 2 /127 R meter e = GV 2 /1.27 R centimeter For B.G., e = GV 2 /1.27 R = 1.676 V 2 /1.27 R centimeter For M.G., e = GV 2 /1.27 R = 1.0 V 2 /1.27 R centimeter For N.G., e = GV 2 /1.27 R = 0.762 V 2 /1.27 R centimeter

33. THE END