Railway Engineering

Permanent way

The combination of rails, fitted on sleepers and resting on

ballast and subgrade is called railway track or permanent way.

Cross section of a Permanent Way

Functions of its components

Rails – act as girders to transmit the wheel load to the sleepers.

Sleepers – hold the rails in ‘position’ wrt level, tilt, and gauge.

Ballast – distributes load over wide area of formation, and holds

sleepers in position.
Note: Refer in detail about types and functions in text book (IMP).

Gauge
It is the clear distance between inner faces or running faces of
two rails.

Standard gauge (BG) = 1.676 m

Meter gauge (MG) = 1.000 m
Narrow gauge (NG) = 0.762 m
Feeder or Light gauge (FG) = 0.610 m
Selection of Gauge

 Cost of construction – wider gauges requires more cost due to

increased land acquisition, earth work, ballast, and rolling stock.
 Volume and nature of traffic – higher volume and heavy loads
demand wider gauges due to use of better traction and
locomotives.
 Physical features of area (terrain) – narrow gauges are warranted
in hilly regions, wider gauges are not possible due to steep
gradients and sharp curves.
 Speed of movement – higher speeds demand wider gauges
because speed is proportional to wheel diameter which in turn
proportional to gauge of the track.
 Coning of wheels
The tread of wheels are tapered at a slope of 1 in 20, this is known
as coning of wheels.

Advantages
 Reduces wear and tear of wheel flanges and rails, which is due
to rubbing action of flanges with inside faces of rail heads.
 To prevent wheels from slipping to some extent.
Adzing of Sleepers
 Geometric Design of Track
Objective

To have safe movement of trains, all the geometric elements of the

track should be designed according to the required design speed.

Geometric elements

 Gradients and grade compensation

 Speed of train

 Radius or degree of the curve

 Cant or superelevation

 Widening of gauge on curves

Gradients and grade compensation

Gradient
It is rate of rise or fall per unit horizontal length.

Expressed in 1 in x or n in 100 or n%

Types

 Ruling gradient
 Momentum gradient
 Pusher or Helper gradient
 Gradients at station yards

Ruling gradient – it is the maximum gradient allowed on the

track, and this value determines maximum load the engine can
haul on the section.
According to Indian Railways

Plain terrain – 1 in 150 to 1 in 200

Hilly terrain – 1 in 100 to 1 in 150

Momentum gradient

In hilly terrain, typically a falling gradient is usually followed by a

rising gradient.

A train while coming down on falling grade attains a

momentum, this momentum in turn gives additional kinetic
energy to the train, this enables the train to overcome gradients
steeper than ruling gradient for a certain length is known as
momentum gradient.
Pusher or Helper gradient

A gradient which is steeper than ruling gradient requires an

additional assisting engine to push train is known as pusher or
helper gradient.

Gradient at station yards

These gradients are designed very low, otherwise the train

moves with a gentle push during maintenance activity.

Grade compensation on curves

If curve lies on the ruling gradient, the resistance due to

curvature adds to the additional load due to grade. In order to
avoid resistance beyond the allowable limits, the gradients are
reduced on the curves and this reduction in gradients is known
as grade compensation on curves.
Grade compensation

According to Indian Railways

For BG - 0.04 % per degree of curve
For MG - 0.03 % per degree of curve
For NG - 0.02 % per degree of curve
Speed of the train

Maximum permissible speed on BG = 160 kmph

MG = 100 kmph

Safe speed on curves

According to Martin’s empirical equation (old practice)

a) Where transition curve exists
For BG and MG, safe speed V = 4.35 𝑅 − 67 or
V = 4.4 𝑅 − 70 kmph
For NG V = 3.6 𝑅 − 6.1 or V = 3.65 𝑅 − 6
subject to a max 50 kmph
b) Where transition curve absent
4
For BG ,MG, and NG V = of the speed calculated in (a)
5
above
Martin’s equation no longer followed in India, according to Indian
Railways
For BG
(𝐶𝑎+𝐶𝑑)×𝑅
V= OR V = 0.27(Ca+Cd)×R
13.76
Where
V – max speed in kmph
Ca – Actual cant in mm
Cd – cant deficiency allowed in mm
R – radius in m

For MG
V = 0.347(Ca+Cd)×R

For NG
V = 3.65 𝑅 − 6
Radius of the curve

We know the safe speed on the curve is given as

For BG and MG, V = 4.35 𝑅 − 67 or V = 4.4 𝑅 − 70 kmph

For NG V = 3.6 𝑅 − 6.1 or V = 3.65 𝑅 − 6

From above equation required radius can be found for a given

speed.
Relation between Radius and Degree of curve

𝐷 3600
=
30 2𝛑𝑅

1720
D=
𝑅

where, R is in m, and D is in degrees.

Superelevation or Cant

To counteract the effect of centrifugal force on horizontal curve,

the outer rail is raised above the inner rail by a certain amount.

The raised elevation of outer rail above inner rail at a horizontal

curve is known as superelevation or cant.

Objective

 To prevent side wear of rails and derailment.

 To provide equal distribution of wheel loads on two rails.

 Superelevation Design
𝑊𝑣2
Centrifugal force F =
𝑔𝑅

Resolve forces parallel to inclined surface

F cosα = W sin α
𝑊𝑣2 𝐺 𝑒
× =W×
𝑔𝑅 𝑆 𝑆

𝑣2
e= × 𝐺 in meters α
𝑔𝑅
F

𝐺(0.278𝑉)2 𝐺𝑉2 𝐺𝑉2 α

= = m= cm outer rail W
𝑔𝑅 127𝑅 1.27𝑅
inner rail
e S
where G is in m, V is in kmph, and R is in m α
G
Therefore,

1.676 𝑉2 𝑉2
For BG, e = = 1.315 cm
1.27 𝑅 𝑅

1.000 𝑉2 𝑉2
For MG, e = = 0.80 cm
1.27 𝑅 𝑅

0.762 𝑉2 𝑉2
For NG, e = = 0.60 cm
1.27 𝑅 𝑅

The above cant is known as equilibrium cant, and is designed for

average speed of all the trains on the curve.

According to Indian Railways, average speed is calculated as

follows:
a) When the max sanctioned speed of the section on BG and MG
is over 50 kmph i.e., Vmax > 50 kmph

3
i) Speed = × Vmax subject to a minimum of 50 kmph
ii) Safe speed on curves given by formula

The average speed is (i) or (ii) whichever is less

b) When the max sanctioned speed of the section on BG and MG is

50 kmph or less i.e., Vmax ≤ 50 kmph

i) Speed = Vmax

ii) Safe speed on curves given by formula

The average speed is (i) or (ii) whichever is less

Limits of Superelevation and Cant-Deficiency

Maximum Superelevation

To take care of slow and fast moving trains on the same curve,
1
maximum superelevation is limited to of the gauge.
10

1
i) Maximum SE for BG = × 1.676 m = 0.1676 m = 16.7 cm
10
1
ii) Maximum SE for MG = × 1.000 m = 0.1000 m = 10.0 cm
10
1
iii) Maximum SE for NG = × 0.762 m = 0.0762 m = 7.6 cm
10

Cant Deficiency

It is the difference between the equilibrium cant required for

maximum permissible speed on a curve and the actual cant
provided on the basis of average speed of trains.
Limits of Cant Deficiency

Gauge Cd for speeds up to 100 Cd for speeds > 100

kmph kmph

BG 7.6 cm 10.0 cm
MG 5.1 cm Not specified
NG 3.8 cm Not specified
 Negative Superelevation

Hence, speed on both tracks must be restricted, particularly on

branch line.
Prob 1.
If a 80 curve track diverges from a main curve of 50 in an opposite
direction in the layout of a BG yard, calculate the superelevation
and the speed on the branch line if the maximum speed
permitted on the main line is 45 kmph.

Sol:

i) Equilibrium cant required for a max speed of 45 kmph on main

track is

𝐺𝑉2
e=
1.27𝑅

1720 1720
R= =
𝐷 5

1.676×45×45 5
= × = 7.78 cm
1.27 1720
ii) Permissible cant deficiency for BG is 7.6 cm

iii) Cant required for main line is = 7.78 - 7.6 = 0.18 cm

iv) ∴ Cant to be provided for branch line = -0.18 cm i.e., negative

cant

v) Cant required for max permissible speed = Ca + Cd

= -0.18 + 7.6

= 7.42 cm

𝐺𝑉2
vi) Hence, permissible speed on branch line, from e = is
1.27𝑅

1.676×𝑉2 8
7.42 = ×
1.27 1720

V = 34.7 kmph.
Prob 2.
A 50 curve diverges from a 30 main curve in reverse direction in
the layout of a BG yard. If the speed on the branch line is
restricted to 35 kmph, determine the restricted speed on the main
line.

Sol:

i) Equilibrium cant required for a max speed of 35 kmph on branch

line is

𝐺𝑉2
e=
1.27𝑅

1720 1720
R= =
𝐷 5

1.676×35×35 5
= × = 4.71 cm
1.27 1720
ii) Permissible cant deficiency for BG is 7.6 cm

iii) Cant required for branch line is = 4.71 - 7.6 = -2.89 cm i.e.,
negative cant

iv) ∴ Cant to be provided for main line = 2.89 cm

v) Cant required for max permissible speed = Ca + Cd

= 2.89 + 7.6

= 10.49 cm

𝐺𝑉2
vi) Hence, permissible speed on main line, from e = is
1.27𝑅

1.676×𝑉2 3
10.49 = ×
1.27 1720

V = 67.4 kmph. Say 65 kmph

 Widening of gauge on curves
Due to rigidity of wheel base, when the outer wheel of front axle
strikes against the outer rail, the outer wheel of rear axle bears a
gap with the outer rail.

An extra widening of outer rail is required, otherwise there is every

possibility of tilting of rails outwards.

The required extra width (d) is given as

2
13 𝐵+𝐿
d=
𝑅

where B = rigid wheel base, for BG is 6 m, for MG is 4.88 m

R = radius of the curve in m
L = 0.02 ℎ2 + 𝐷. ℎ in cm
h = depth of wheel flange below rail top level in cm
D = diameter of wheel in cm
 Shift
When a transition curve is to be fitted in between the straight and
circular track; the original circular curve is to be shifted inwards, this
is known as shift of circular curve.

𝐿2
Shift S =
24𝑅

where S = shift in m
L = length of transition curve in m
R = radius of circular curve in m
 Points and Crossings
These are the arrangements made in railway tracks to change train
from one track to another.

Typical layout of left hand turnout

Components of Turnout

 A pair of points or switches (ABCD and EFPQ)

 A vee crossing (GHIJ)
 Two check rails
 Four lead rails

Important terms used in Points and Crossings

 Facing direction
 Trailing direction
 Right hand and Left hand turnouts
 Right hand and Left hand switches
Line diagram of right hand turnout

Line diagram of left hand turnout