1. Introduction To Railway Engineering
Mr. Vedprakash Maralapalle, Asst. Professor
Department: B.E. Civil Engineering
Subject: Transportation Engineering
Semester: VI
Teaching Aids Service by KRRC Information Section

2. Contents Gauges Railway track Cross section Coning of wheel
Components
Functions
Cross section
Coning of wheel
Creep of rail
Failure and joints
Ballast
Sleeper
Drainage

3. Rail gauge:
In rail transport, track gauge is the spacing of the rails on a railway track and is measured between the inner faces of the load-bearing rails.
All vehicles on a network must have running gear that is compatible with the track gauge,
Indian Railways uses four gauges:
1.676 m (5 ft 6 in) Broad Gauge (BG), which is also called Indian gauge,
1.000 m (3 ft 3 3⁄8 in) metre gauge (MG),
.762 m (2 ft 6 in) Narrow Gauge (NG), and
.610 m (2 ft) Narrow Gauge.

4. Railway track Also known as the permanent way
Consisting of the rails, fasteners, railroad ties (sleepers) and ballast (or slab track), plus the underlying subgrade

5. COMPONENT PARTS OF A PERMANENT WAY :-
The following are the component parts of a permanent way or a railway track :-
Formation or sub grade ;
Ballast ;
Sleepers ;
Rails ;
Fixtures and fastenings.

6. REQUIREMENTS OF AN IDEAL PERMANENT WAY
The following are the principal requirements of an ideal permanent way or of a good railway track :-
The gauge of the permanent way should be uniform, correct and it should not get altered.
Both the rails should be at the same level on tangent (straight) portion of the track.
Proper amount of *superelevation should

7. be provided to the outer rail above the inner rail on curved portion of the track.
The permanent way should be sufficiently strong against lateral forces.
The curves, provided in the track, should be properly designed.
An even and uniform gradient should be provided through out the length of the track.
The **tractive resistance of the track should be minimum.
The design of the permanent way should be such that the load of the train is

8. uniformly distributed on both the rails so as to prevent unequal settlement of the track.
It Should provide adequate elasticity in order to prevent the harshness of impacts between the rails and the moving wheel loads of a train.
It should be free from excessive rail joints and all the joining should be properly designed and constructed.
All the components parts such as rails, sleepers, ballast, fixtures and

9. fastenings, etc. should satisfy the design requirements.
All the fixtures and fastenings such as chairs, bearing plates, fish plates, fish bolts, spikes etc. should be strong enough to withstand the stresses occurring in the track.
All the *points and crossings, laid in the permanent way, should be properly designed and carefully constructed.
It should be provided with fence near

10. level crossings and also in urban areas.
It should be provided with proper drainage facilities so as to drain off the rain water quickly away from the track.
It should be provided with safe and strong bridges coming in the alignment of the track.
It should be so constructed that

11. Sleepers
A railroad tie/railway tie/crosstie (North America), or railway sleeper (Europe, Australia & Asia) is a rectangular support for the rails in railroad tracks

12. Ballast
Used to bear the load from the railroad ties, to facilitate drainage of water, and also to keep down vegetation that might interfere with the track structure.
Serves to hold the track in place as the trains roll by. It is typically made of crushed stone, although ballast has sometimes consisted of other, less suitable materials.
Thickness should not be less than 150mm
For high speed railway lines, it can be min. 1.5m

13. Rail cross-section

14. Coning of wheel
Flanges of the wheels are never made flat but are in the shape of a cone with a slope of 1 to 20, called coning of wheel.
Purpose:
To keep the train in its central position of the rails, coning does not allow any sidewise movement on a straight track.
To allow the wheels to move different distances on a curved track and thereby reduce wear and tear.

15. Advantage Disadvantage: Maintains the gauge properly
The wear at the head of rail is uniform
It increases the life of sleepers and the rails.
Disadvantage:
With no base plate, sleepers in the outer edge of rail is damaged
Gauge is widened sometimes
Pr. Of the horizontal component near inner edge has a tendency to wear the rail quickly.

16. Track stresses Caused due to: Wheel loads Hammer blow
(due to revolving masses like crank pins, connecting rods, coupled rods, bolts)
Horizontal thrust
Pr. Exerted by flanges of wheel
Stresses due to irregularity in tracks

17. Stress in sleeper
Subjected to permanent stress arising from the fastenings.
Stress in sleeper depend upon:
Wheel load
Weight transfer from wheel to wheel
Speed
Dynamic effect of wheels on rails
Elasticity of the rail
Efficiency of fastenings

18. Stress in ballast

19. Wear on rails
Axle loads are abnormally high and the train moves with very fast speed, then the connected exceeds the elastic limit resulting in metal flow.
Classification of wear:
Wear on head of rail
Wear on ends of rail
Wear of rail on curve

20. WEAR ON HEAD OF RAIL
Wear on head of rail is due to abrasion on moving rails.
Due to grinding action of sand or dust between the rails and wheels of the train.
When train starts or applies brakes, the wheel just slides on the rails causing wear on the head.
Load coming on to a track may exceed the carrying capacity of the section. Thus causing the wear in the head of rail.

21. Wear on ends of rail
The ballast under the sleepers will loosen due to increase in the intensity of vibrations, also he sleeper will depressed due the displacement of ballast, also the fish plates will get loose under the constant impact of increasing vibrations

22. Rail failure Abrupt change of section of rail
Notches with corners in the foot of the rails
Failure occurs because of:
Crushed heads
Split heads
Square or angular break
Split web
Horizontal fissure

23. CREEP OF RAILS Definition
It is a horizontal movement of rails in a track. It can be minimized but cannot be stopped.

24. Causes Of Creep There are three main causes of Creep
Wave motion of trains.
Expansion and contraction of rails due to variation in temperature.
Due to starting, accelerating, slowing down (decelerating) and stopping of trains.

25. Direction Of Creep Alignment Of Track:
Creep is more on curve track than on a tangent portion (straight track).
Grades:
In upgrades tracks, creep will be less and in down grades track creep will be more.
Direction of movement of trains:
Creep will be more in the direction to which the loaded train moves more.

26. Extent Of Creep
Creep does not vary at some constant rate. (it is not constant)
Creep does not continue in one direction only.
Creep for two rails of the track will not be in equal amount.
Result Of Creep
Expansion gap is reduced, buckling of track take place.
Sleepers are moved out of a square.
Crossing points get disturbed.

27. TYPES OF CROSSING

28. TYPES OF CROSSING Square Crossing Diamond Crossing Cross Over
Scissor Crossing
Symmetrical Split

29. SQUARE CROSSING
When two railway lines cross each other at 90o it is called Square Crossing

30. DIAMOND CROSSING
Angle of intersection (crossing angle) of two tracks is when not 900 , then crossing is called diamond crossing

31. CROSS OVER
A cross over is introduced to transfer a train from one track to another track which may or may not be parallel to each other

32. SCISSOR CROSSING
If two cross overs are required between two parallel tracks and there is no sufficient space for crossing to be kept separate, then they are made to over-lap each other and result is a scissor crossing.

33. SYMMETRICAL SPLIT
If radius of main track is equal to the radius of turn out curve, then the turn out is known as symmetrical split.

34. Rail joints
Strength of rail joint is only 50% of the strength of a rail
Requirements of an ideal rail joint:
Two rail ends should remain true in line both laterally and vertically when train moves in the track
Rail joint should be as strong as stiff as rail itself
Should provide enough space for expansion
Should be easily disconnect able
Should not allow rail ends to get battered.

35. Types of rail joints Supported rail joint Suspended rail joint
Bridge joint
Base joint
Welded rail joint
Staggered rail joint
Square joint
Compromise joint

36. Supported rail joint Bridge joint Suspended rail joint
Compromise joint
Bridge joint

37. Geometric Design Alignment Gradient and grade compensation
Super elevation
Equilibrium cant and cant deficiency
Horizontal curves
Transition curves
Signaling and interlocking
Traction and tractive resistance

38. Rail alignment
The horizontal alignment is done by using a predefined length of string line to measure along the gauge side of the reference rail. It is the distance (in inches or millimeters) from the midpoint of the string line to the gauge of the reference rail.
The design horizontal alignment on the curved track in the United States is 1 inch for each degree of curvature. Any other readings indicate deviations.
The vertical alignment is the surface uniformity in the vertical plane. The measurement of uniformity is done using a predefined length of string line (normally the same length used in horizontal alignment) along the track. If the midpoint of the measurement has higher elevation, it is called hump deviation. On the other hand, if the midpoint has lower elevation, it is called dip deviation

39. Gradients
Gradients limit the load that a locomotive can haul, including the weight of the locomotive itself. On a 1% gradient (1 in 100) a locomotive can pull half (or less) of the load that it can pull on level track.
Extremely steep gradients require the use of cables or some kind of rack railway (such as the Pilatus railway in Switzerland, with a maximum grade of 48% (26°), claimed to be the world's steepest rack railway) to help the train ascend or descend.

40. Grade Compensation
Grade compensation in curvature is provided to off-load the extra resistance due to curvature.
Compensation in BG is 0.04% per degree of curvature. So while providing/fixing the gradient in a curvature, care is to be taken for compensation.
It should be in such a way that calculated grade in curvature without compensation should not be steeper than rulling gradient. 
Get the rulling gradient of the section. As per degree of curvature, calculate the amount of compensation. Flatten the rulling gradient with compensation. Say the gradient after compensation is (C). Check out proposed gradient in curvature should be equal or flatter than (C). 
Thus in a section having ruling gradient of 1 in 150, if a 3 degree curvature is provided, then the compensated gradient will be [(1/150)*100-(0.04*3)] = % = 1 in 183. 

41. Super-elevation
The cant of a railway track (also referred to as superelevation) is the difference in elevation (height) between the two edges.
This is normally done where the railway or road is curved; raising the outer rail or the outer edge of the road providing a banked turn, thus allowing vehicles to maneuver through the curve at higher speeds than would otherwise be possible if the surface was flat or level.

42. Function of cant The main functions of cant are to:
Better distribute load across both rails
Reduce rail- and wheel-wear
Neutralize the effect of lateral forces
Improve passenger comfort
Positive cant
Negative cant

43. Cant deficiency
Defined in the context of travel of a rail vehicle at constant speed on a constant radius curve. 
Cant itself is a British synonym for the super elevation of the curve, that is, the elevation of the outside rail minus the elevation of the inside rail.
Cant deficiency is present when a vehicle's speed on a curve is greater than the speed at which the components of wheel to rail force normal to the plane of the track would be the same in aggregate for the outside rails as for the inside rails.
For a constant speed of a running train, the amount of required cant to achieve the balance is called equilibrium cant

45. Traction and Tractive resistance
Definition:
The source through which the locomotive drives power is called traction.
Sources:
Steam
Diesel fuel
Electric supply (AC/DC)

46. Traction Traction has a bearing upon: Load carrying capacity Speed
Economy
Efficiency of service

47. Resistance due to traction
Train resistances (due to rolling stock)
Resistance due to track profile
Traction loss (due to starting and acceleration)
Wind resistances

48. Train resistances Resistances independent of speed are caused due to:
Friction imposed due to train components (locomotive, wagons / compartments), known as journal friction.
Dependent upon type of bearing, lubricant used and temperature of atmosphere.
For roller bearing – 0.5kg to 1.0kg per ton
For coupled boxes – 1.3 to 1.5kg per ton

49. Contd. Resistances independent of speed (Rt1) are caused due to:
Friction between steel wheels and steel rails
Track resistance – wave action of rails
Resistance due to internal parts, e.g. cylinder and rim of driving wheels, etc.
Computation:
Rt1 = w
Where w is weight of train in tons.

50. Contd. Resistance dependent on speed (Rt2): causes due to –
Track irregularities
Vertical movement of wheels on rails
(improper joints and maintenance)
Flange action (oscillations, sways, etc.)

51. Resistance dependent upon speed (Rt2):
Rt2 = wv
Where ‘w’ is weight of train in tons and
‘v’ is speed of train in kmph.

52. Atmospheric resistance (Rt3):
On sides and end of wagons / locomotives
Wind is assumed as not blowing
Computed as:
Rt3 = wv2
Where ‘w’ is weight of train in tons and
‘v’ is speed of train in kmph

53. Train resistance (RT1)= Rt1 + Rt2 + Rt3
= w wv wv2
Resistance due to track profile (RT2):
Caused due to –
Gradients and
Curves

54. Resistance due to gradient (Rg)= w tanθ
Resistance due to curves (Rc):
Factors controlling are:
Rigidity of wheel base
Wear on inner side of outer rail due to flange of leading axle and inner side of inner rail due to flange of trailing axle, causing mount on rail
Extra super-elevation
Slippage of wheel
Poor maintenance of track and components

55. Distance travelled by outer wheel = D1
Distance travelled by inner wheel = D2
Extra distance travelled = D1 - D2
If F is the force of sliding friction,
Then work done = (D1 – D2)F
Mean resistance = F.G / R
Therefore, resistance gets affected by force of sliding friction, gauge of track and degree of curvature.
Recommended values of curve resistances:
Broad gauge, Rc = wD
Meter gauge, Rc = wD
Narrow gauge, Rc = wD

56. Contd. Resistance due to tractive effort (RSA): Get induced due to –
Starting operation
Varies according to the type of an object
For locomotives Rs = 0.15w1
For vehicles Rs = 0.005w2
Where w1 and w2 are weight of locomotive and vehicle respectively
Acceleration given to a locomotive
RA = 0.028w (v2 – v1)/t
Where v1 and v2 are velocity at starting and at the end
T is the time taken from v2 to v1
w is total weight of train

57. Wind resistances Resistance due to wind (RW) Depends upon –
Direction of wind w.r.t movement of train
Wind velocity
Sectional area exposed to wind
RW = AV2
Where A = exposed area in sq. meters
V = velocity of wind in kmph
So, total tractive resistance
RT = RT1 + RT2 + RSA + RW

58. Hauling capacity
Defined as the total load that can be handled by the locomotive. It is an indicative of power available to a locomotive.
Computed as a product of coefficient of friction and weight on the driving wheels.
At the minimum level it should be equal to traction resistances.

59. Factors affecting H.Capacity
Weight coming on the driving wheels
Coefficient of friction
Largely depends on condition of surface
Speed of locomotive
Hauling capacity = μwn = μW
n is the no. of pairs of driving wheels
W is the total load on driving wheels
Maximum axle load in India,
BG = 28.56t
MG = 17.34t
NG = 13.26t

60. Problem
Calculate the maximum permissible train load that can be pulled by a locomotive having four pairs of driving wheels carrying an axle load of 30tonnes each. The train has to run at a speed of 70kmph on a straight level track. Also calculate the reduction in speed, if train has to climb a gradient of 1 in 150. If train climbs the gradient with a 30 curve, then what would be reduction in speed?

61. Solution: Hauling capacity = 1/6th times the load on driving wheels
Therefore hauling capacity = (1/6)*4*30 = 20tonnes
On a straight level track, train resistance is,
= w w*V w*V2
So, 20 = w w* w*702
Solve for w

62. In case the train has to move up the gradient 1 in 200 (0.5 percent)
Total train resistance = w w*V w*V 𝑤∗
Use ‘w’ from previous calculation
Solve V1
If train climbs a gradient with a curve of 30, then additional term will be = w*3
So, 20 = w w*V w*V22 + 𝑤∗ w*3
Find V2
Hence reduction in speed = (70 – V2)kmph

63. Equilibrium super-elevation
e = GV2 / 127R
Where G is the gauge, V is the speed in kmph and R is the radius of curve
G = gauge length + width of rail head
For BG track, G = 1750mm
For MG track, G= 1058mm

64. Thumb rule for S.E calculation
A field engineer can adopt the following thumb rules for determining the super elevation of any curve.
Superelevationfor BG in cm
= (speed in km/h)2 𝟏𝟎𝟎 * 𝐝𝐞𝐠𝐫𝐞𝐞 𝐨𝐟 𝐜𝐮𝐫𝐯𝐞 𝟏𝟑
(b) For M.G track, 3/5th of the earlier value
Problem:
if the maximum sanctioned speed (MSS) of the section is 100 km/h, the equilibrium speed may be taken as 75% of the MSS, i.e.,75 km/h. The super elevation for a 1° curve as calculated by the thumb rule

65. Negative super elevation
When the main line lies on a curve and has a turnout of contrary flexure leading to a branch line, the super elevation necessary for the average speed of trains running over the main line curve cannot be provided.

66. The provision of negative superelevation for the branch line and the reduction in speed over the main line can be calculated as follows.
(i) The equilibrium super elevation for the branch line curve is first calculated using the formula,
e = GV2 / 127R
(ii) The equilibrium super elevation e is reduced by the permissible cant deficiency Cd and the resultant super elevation to be provided is
x = e – Cd
where, x is the super elevation, e is the equilibrium super elevation, and Cd is 75 mm for BG and 50 mm for MG.
The value of Cd is generally higher than that of e, and, therefore, x is normally negative. The branch line thus has a negative super elevation of x.

67. Problem
Calculate the super elevation and the maximum permissible speed for a 2° BG transitioned curve on a high-speed route with a maximum sanctioned speed of 110 km/h. The speed for calculating the equilibrium super elevation as decided by the chief engineer is 80 km/h and the booked speed of goods trains is 50 km/h.

68. Solution
Find R = 1750 / D
Super elevation for equilibrium speed = GV2 / 127R
S.E for max. sanctioned speed 110km/hr
Find cant deficiency
S.E for booked speed of 50km/hr
Find out cant excess (if it is less than specified value, then permissible)
Max. speed potential =
V = √ 𝑐 𝑎 + 𝑐 𝑑 ∗𝑅 13.76
Where ca is the maximum of cant excess and S.E for equilibrium speed

69. Problem
A BG branch line track takes off as a contrary flexure through a 1 in 12 turnout from a main line track of a 3° curvature. Due to the turnout, the maximum permissible speed on the branch line is 30 km/h. Calculate the negative super elevation to be provided on the branch line track and the maximum permissible speed on the main line track (when it takes off from a straight track).

70. Solution For a branch line track, degree of curve = 4-3= 10
Find out radius of curve for branch line
Find S.E
Round it off to the higher multiple of 5
Find negative super elevation for the branch line,
x = e – cd
for main line, S.E is same as of branch line but with +ve sign
For equilibrium speed, e = x + cd
Solve for equilibrium speed

71. Railway drainage
Signaling and interlocking