Railway
Railway
Railway Track
Chapter 1
1.1 INTRODUCTION
y The concept of railway is the movement of train wagons or passenger
bogies fitted with steel wheels running over two parallel steel rails of
the railway track.
y The main advantage of the railway is that it can transport many
passengers or large quantities of goods at a time.
y Rail transportation is economical.
y Indian Railways was introduced in 1853. The Network of Indian railway is
one of the largest in the world.
Definitions
1
Chapter 1
1.3 RAILS
Definitions
Geometric Design of Railway Track
Rails: Steel girder which provides a smooth and hard surface for
movement of locomotive and wagons are called rails.
2
Chapter 1
y The material used in rails should be such that it offers minimum wear
so that frequent replacement and wearing failure of rails is avoided.
y Rails transmit the loads to sleepers and consequently reduce pressure
on the ballast and formation below.
Head
F ishing
Head
angle
Foot
E
1:3
side
1:3
D
Neutral Horizontal axis Geometric Design of Railway Track
A
Vertical axis
Height
Tension
side
1 :6 1:6
F
3
Chapter 1
y Selection of rails
Since weight per unit length is the vital parameter in the design of rail.
The following factors govern the suitability of weight of rail:
a) Speed of the train
b) The gauge of the track
c) The axle load and nature of traffic
d) Type of rails
e) Spacing of sleepers (sleeper density)
f) Maximum permissible wear on top of rails (5% of the weight of rail
is allowed)
y Kinks in rails
When the end of adjoining rails moves slightly out of position, ‘shoulder‘
or ‘kinks’ are formed.
Kinks formation
y Length of Rails
The rails of larger length are preferred because they are economical and
provide more strength. The joint between two rails is the weakest point
of a track. For fewer joints, the less number of fish plates are required,
which means minimal maintenance cost along with the smoother and
comfortable ride.
On Indian Railways, the standard lengths are:
12.80 m (42 ft) for B.G. (say 13 m)
11.89 m (39 ft) for M.G. (say 12 m)
Geometric Design of Railway Track
y Hogged rails
Definitions
Hogged rails: The rails which get deflected due to the battering action
of wheel over the ends of the rails are called hogged rails.
4
Chapter 1
y The hogging at the end of the rail is due to loose packing under the
joints or a loose fish plate.
Definitions
5
Chapter 1
Creep Formation
6
Chapter 1
iii) To build up worn-out points and rails on the sharp curves.
iv) To build up the burnt portion of the railhead, which is caused due to
slippage of wheels over the rails or other defects or spots in rail steel.
2) Advantages of welding rails:
y Welding satisfies the condition of the perfect joint and hence
increases the life of the rail.
y There is a reduction in the maintenance cost of the track by about
20 to 40 percent.
y It reduces the creep due to an increase in the length of the rail and,
in turn, friction as well.
y Expansion effect due to temperature is reduced, which in turn also
reduces the creep.
y Due to discontinuity of joints, a source of track weakness is reduced.
The defects, such as hammering at rail joints, displacement of joints,
disturbance in alignment and running surface, which result in bad
riding quality, are eliminated.
y Long rail lengths being heavier, dampen the intensity of high
frequency vibration due to moving loads.
y Welding increased the life of rails due to a decrease in the wear of
rails at joints.
y Welding facilitates track circuiting on electrified tracks.
y Welded rails provided on large bridges for the span length are helpful
as they result in better performance.
y Welded rails provision on curves is under investigation. However,
maximum curve length may be welded depending upon the resistance
and lateral displacement of the track.
3) Length of welded rails:
y Joints in the rail track are the weakest part.
y The lesser the number of joints lesser is the cost of maintenance
required.
y So, the length of rails should be as long as possible, but the
limitations are:
i) Facilities available for rail manufacturing at a reasonable cost. Geometric Design of Railway Track
ii) Length of rail that can be transported using the longest wagon
available.
iii) Limitations arising due to the availability of machinery for
handing and shifting of long rails.
4) Welded rails:
y Increase in length of rail due to expansion
dl = l × a × Dt
l = length of rail
a = Coefficient of expansion in per °C
Dt = Rise in temperature above the temperature at the time of
construction
7
Chapter 1
y The rail fittings have a tendency to hold down the rails to the sleeper,
hence, restricting their movement and transferring the longitudinal
stress in the transverse and vertical direction.
y These stresses way pull the alignment of the track due to buckling
and may result in derailment of trains.
Consider:
E = Modulus of elasticity of steel (kg/cm2)
A = Cross-sectional area of rail (cm2)
F = Force in kg, required to prevent expansion due to change in
temperature.
F.l = dl.A.E
Fl = l × a × Dt.A.E
F = α∆tAE
y This is referred ‘Locking up of longitudinal thermal stresses’ and has
proved that longitudinal movement of rails takes place only at the
ends (known as breathing length) due to temperature variation and
the absence of the resisting force of track while the central portion
of the rails remain fixed (known as fixed length) due to resistance
offered by rails by means of sleeper, rail fastening and ballast against
any expansion due to change in temperature.
Example 1.1: If the temperature rise is 29°C, then the increase in the length
of rail of 12.8 m will be _____. [Given: a= 1.18 × 10–5 per °C]
a) 4.38 mm b) 3.97 mm
c) 4.12 mm d) 5.83 mm
Sol:
Increase in length (dl) = l × a × Dt
= 12.8 × 1.18 × 10–5 × 29
= 4.38 × 10–3 m = 4.38 mm
Correct answer is a).
Example 1.2: Find the length of track
Geometric Design of Railway Track
8
Chapter 1
Sol:
Force required to prevent the expansion due to change in temperature
F = atAE
F = 1.16 × 10–5 × 30 × 60 × 21.45 × 105
F = 44787.6 kg
a) Length of track to overcome temperature stress
F
(Lt) =
resistance of track
44787.6
Lt =
700
Lt = 63.98 m
b) Length of welded track to prevent creep for equilibrium
= 2Lt
= 2 × 63.98
= 127.96 m
1.4 SLEEPERS
1) Function of sleepers:
y Hold the rail to the correct gauge.
y To hold the rail at the proper level, i.e., at turnouts and cross-overs.
y To act as an elastic medium in between the ballast and rails to
absorb the blows and vibration of moving loads.
y To distribute the load from the rails to the underlying ballast or to
the girders in case of bridges.
Definitions
9
Chapter 1
2) Types of sleepers:
10
Chapter 1
S.No. Point of Wooden Sleeper Cast Iron Steel Sleeper Concrete
Comparison Sleeper Sleeper
7) Track fittings Requires less Required Requires less Requires less
fittings more fittings fittings fittings
8) Elasticity Good Not so good Not so good Not good
9) Laying and Easiest Difficult Easy due to Difficult
Relaying due to large light weight. by manual
number of labour. Easy
fittings if mechanical
devices are
used
10) Rigidity of Poor both Better than Better than Best
track laterally and timber timber because of
longitudinally sleepers sleepers heavy dead
weight
11) Suitability of Generally Suitable only Suitable only Suitable for
track suitable in for stone for stone any location
all locations ballast. ballast. on railway
except areas Unsuitable in Unsuitable track.
of vermins station yards for station
and white yards and
ants. Specially coastal areas
suitable for
points, bridges,
station yards
and level
crossings.
12) Track Best Restricted, Restricted Moderate
circuiting insulating insulating
Geometric Design of Railway Track
pads are pads are
necessary necessary
13) Scrap value Very little Highest Next to C.I. Nil
14) Gauge Does not Slight shift Maintains Depends
maintain proper in gauge due proper gauge upon design,
gauge to play in improved
the bar and design
socket. maintains
proper
gauge.
11
Chapter 1
r
y The number of sleepers per rail varies in India from M+4 to M+7 for
the main track, here M is length of rail in meters.
y Length of rail for B.G track is 12.8 m approx. 13 m.
Example 1.3: Which of the following is the expression for sleeper density for
Geometric Design of Railway Track
12
Chapter 1
Example 1.4: Using sleeper density of M + 6 find out the number of railway
sleepers required for a track of 648 meters in length. (BG Track).
Sol:
Length of each rail on a B.G track = 12.8 m ~ 13 m
Total number of rails required will be
648
=
13
= 49.846 ~ 50
Sleeper density = M + 6
Number of sleeper under each rail = 13 + 6 = 19
Total number of sleeper = 50 × 19 = 950 sleepers
13
Chapter 1
1.6 BALLAST
Definitions
Ballast: Granular material is placed and packed below and around the
sleepers. This granular material transmits wheel load from sleepers to
the formation. Ballast also helps in drainage in the track.
P P
Sleeper spacing (S)
45° 45°
Db Db
Definitions
y The permanent way consists of a series of rails joined by fishplates and Geometric Design of Railway Track
bolts.
y Rails are then fixed to sleepers by different types of fastenings.
y The rails act as girders to transmit the wheel load to the sleeper.
y The rails are fitted in a specific position considering the tilt, gauge and
level.
y Sleepers transmit the load from rails to the ballast.
y The ballast holds the sleeper in position and distributes the load over
the formation.
y The sleepers are properly placed, packed and boxed with ballast.
15
Chapter 1
Ballast shoulder
Gauge
Ballast Cess
Rails
Sleeper
Trolley reuse
1 .5
:1
2.5m
Ballast cushion
1 .5
:1
Sub-ballast
:1
2
2
:1
G.L. of murum G.L.
Ballast base
Formation width
16
Chapter 1
xi) The components of the track, i.e., the rails, fittings, sleepers, ballast
and formation should fulfil their intended purpose.
xii) There should be adequate provision for easy renewals and
replacement.
xiii) The track structure should be strong with low initial and
maintenance cost.
3.4m
2.7m
1.676 m
12.7cm
20.3cm
1.8 1.8
m m
2
:1
:1
d 30.5cm
2
Turfed 4.4m
5.5m 2d 6.1 m 2d 5.5m
Temporary
land for
Permanent land narrow
(17.1 + 4 × Embankment depth in meters) pits
Fig. 1.5 The Cross-Section of a B.G. Track in Embankment (on Straight Track)
Permanent land
(21.6 + 3 × Depth of cutting + spoil bank)
Turfed
Fig. 1.6 The Cross-Section of a B.G. Track in Cutting
for Double Line (on Straight Track)
17
Chapter 1
Flanges of
Slope 1 in 20 wheels Slope 1 in 20
Wheel Axle Wheel
diameter = D diameter = D
Slope 1 in 20 G Slope 1 in 20
B
Rails
Slope 1 in 20 Slope 1 in 20
Sleeper
:1
1.5
1.5
Ballast
:1
Geometric Design of Railway Track
18
Chapter 1
Cause of Derailment
1.9.1 Gradient:
y Any departure of the track from the level surface is known as grade or
gradient.
y Reasons for providing gradient are:
i) To give a uniform rate of rise and fall
ii) To reach various stations located at different locations.
iii) To reduce the cost of construction
g
u t
Geometric Design of Railway Track
g g g s y
a) Ruling gradient:
� The gradient which helps in determining the maximum load that the
engine can haul on the rail section is known as the ruling gradient.
� While determining the ruling gradient of a section, it will not only
be the severity of the gradient that will come into play but also the
length of the gradient and its position.
19
Chapter 1
b) Momentum gradient:
� The gradient, which is steeper than the ruling gradient, does not
determine the maximum load of the train but is used to find a
suitable position on the track. The train, before reaching these
gradients, it obtains sufficient momentum to negotiate them, which
are referred to as momentum gradient.
� The rising gradient is called the momentum gradient, and in some
cases, a steeper gradient more than the ruling grade can be adopted.
20
Chapter 1
y In India, compensation for curvature is given at 0.04% per degree of
curve for B.G., 0.03% for M.G. and 0.02% for N.G. In terms of radius of
curves in meters 4 is 70/R for B.G., 52..5/R for M.G. and 35/R for N.G.
w
v
21
Chapter 1
The empirical formula given by Martin’s for the calculation of safe speed
on curves are no longer followed by Indian Railways.
The maximum speed for the transition curve is now determined in
Indian Railway as per the revised formula given as:
For BG track
V=
(Ca + Cd ) R
13.76
V = Maximum speed in kmph
Ca = Actual cant (superelevation) mm
Cd = Cant deficiency permitted in mm
R = Radius in meters
Note:
The above equation for maximum speed for BG tracks is derived from the
GV2
formula of equilibrium superelevation i.e. e = on assumption that G
127R
for BG is 1750 mm (Centre of Centre distance between rails)
The maximum cant deficiency value for different gauges for Indian Railway
are:
1) BG 75 mm
2) MG 50 mm
Geometric Design of Railway Track
3) NG 40 mm
Table 1.3 Cant Deficiency
22
Chapter 1
30 m
Circular Curve
R
D°
1720
D= (R is the radius in metre)
R
y The smallest radius and the largest degree for the curvature are
restricted on the basis of:
i) Wheel base of the vehicle:
There is a chance of derailment when the degree of curve is large
for the length of the wheel base, resisting the free manoeuvre of
the vehicle along the curve.
23
Chapter 1
24
Chapter 1
i) To introduce the centripetal force for counteracting the effect of
centrifugal force, which enables the faster movement of trains on
curves. This will also prevent derailment and reduce the side wear
and creep of rails.
ii) To provide equal distribution of wheel loads on two rails so that
there is no tendency for the track to move out of position due
to more load on the outer rail. This reduces the wear of rails,
equipment and results in saving in maintenance costs.
iii) For the safe movement of goods and comfortable ride to passengers
by providing a smooth and even track.
Relationship of superelevation (e), with gauge (G), speed (V) and
radius of the curve (R). Using the following notations,
W = Weight of moving vehicle in kg.
n = Speed of vehicle in m/sec.
V = Speed of vehicle in kmph
R = Radius of curve in meters.
G = Gauge of track in meters.
g = Acceleration due to gravity in m/sec2 Geometric Design of Railway Track
a = Angle oof inclination.
S = Length of inclined surface in metres.
Centrifugal force is given by
Wv 2
F = …(1)
gR
Resolving the forces along the inclined surface, we get
F cos a = W sin a …(2)
Wv 2 G
where F = , cos α =
gR S
25
Chapter 1
e
and sin a =
S
Therefore, equation 2) becomes
Wv 2 G e
× = W×
gR S S
v2
Therefore, e = × G metres …(3)
gR
Where v is in m/sec.
( )
2
G 0.278V
= m. where V is in kmph
9.81R
GV2
= m. …(4)
127R
GV2
= cm. …(5)
1.27 R
Where G is in metres.
V is in kmph.
R is in metres.
In India G for B.G. = 1.676 m
M.G. = 1.0 m
and N.G. = 0.762 m
1.676V2 V2
so, for B.G., e = = 1.315 cm
1.27R R
y The cant or superelevation obtained from equations 4) and 5) is known
as equilibrium cant.
y The cant is said to be in equilibrium when the lateral force and wheel
loads are almost equal. This equilibrium cant is provided on the basis
of the average speed of the train.
y A superelevation is to be provided such that faster trains pass safely
Geometric Design of Railway Track
26
Chapter 1
1.9.6 Limit of superelevation and cant-deficiency
y Superelevation should be such that the track is suitable for various
trains running at different speeds.
y There are limits to the amount of superelevation which may be provided
safely.
y Normally, the maximum superelevation value permissible to railway
board is 1/10th of gauge.
y The maximum permissible value in India for different gauge are:
27
Chapter 1
134 × L
Vmax =
D
where L = length of transition curve based on the rate of change of cant as
38 mm/sec for normal speed and 55 mm/sec for high speeds.
b) For high speeds above 100 kmph:
Geometric Design of Railway Track
198 × L
Vmax =
e
198 × L
Vmax =
D
1.9.8 Negative superelevation:
y When the curved mainline has a turnout on the opposite side leading
to a branch line, the required superelevation for the average speeds of
trains running over the mainline cannot be provided.
28
Chapter 1
Outer rail
Outer
S.E. rail O S.E.
S
Inner R P
Inner
Y
rail X rail
Bra
Y
nc
k
ac
ht
tr
ra c
in
Ma
k
Crossing
M N Points
y From the above figure, MO, which is the outer rail of the mainline curve
must be higher than inner rail NP or point M should be higher than point N.
y For the branch line, however, NS should be higher than MR or the point
N should be higher than point M. These two contradictory conditions
cannot be met at the same time within are layout.
y So, instead of the outer rail NS on the branch line being higher, it is kept
lower than the inner rail MR.
y In such cases, the branch line curve has a negative superelevation and
therefore speed on both tracks must be restricted, particularly on a
branch line.
y The methodology of working out the speed on main line, branch line,
and negative superelevation on branch line are: Geometric Design of Railway Track
a) The equilibrium superelevation or cant on branch line is calculated
GV2
by formula, e = , after assuming a speed on branch line.
127R
b) The permissible cant deficiency is deducted from the equilibrium
cant as obtained in step a).
c) The difference obtained (Equilibrium cant-permissible cant
deficiency) will give the negative superelevation to be used on the
branch line.
29
Chapter 1
Sol:
Degree of curve = 6°
GV2
eactual =
127R
G = 1 m for MG
V = 50 km/hr
1720
R=
6
1 × 502
eacutal =
1720
127 ×
6
eactual = 0.06867 m or 6.87 cm
for speed <100 kmph, for MG limit of cant deficiency is equal to 5.1 cm
eth = ea + ed
= 6.87 + 5.1
eth = 11.97 cm
2
GVmax
so, eth =
127R
Geometric Design of Railway Track
2
1 × Vmax
0.1197 =
1720
127 ×
6
Vmax = 66 kmph
Checking by Martin’s formula
Vmax = 4.35 R –67
30
Chapter 1
1720
= 4.35 –67
6
Vmax = 64.47 kmph
Therefore, maximum permissible speed Vmax = 64.47 kmph
31
Chapter 1
Definitions
t c
c p
Bernoulli’s
is
ax
Leminiscate
or
aj
M
Spiral
Geometric Design of Railway Track
45° Cubic
45° parabola
φ = 3α
0
Fig. 1.11 Types of Transition Curves
32
Chapter 1
1.10.3 Length of transition curve
y The length of the transition curve is the length along the centre line of
the track from the intersection of the straight to the circular curve.
y Half of the transition curve is provided in the straight and the other half
in the circular curve.
y Indian Railway specifies that length of transition curve should be
maximum of the following:
i) L = 7.2e
Where e = actual super-elevation in centimeters
ii) L = 0.073D × Vmax
(Based on rate of change of cant deficiency)
iii) L = 0.073e × Vmax
(Based on rate of change of superelevation)
Here,
L = Length of transition curve
E = Actual cant or superelevation in cm
D = Cant deficiency for maximum speed in cm
V = Maximum speed in kmph
Shifted
u rve
circular c Original
L y ∆ circular curve
X= L/2
L/2 C .S.
S.C.
P.C. P.T. S.T
P.S . Shift S(Shift)
or .
or T T1
R–S
R
Example 1.8: What is the length of transition curve for a B.G. curved track having
5° curvature and a cant of 11 cm ? The maximum permissible speed on
curve is 90 kmph.
33
Chapter 1
Sol:
Length of curve is maximum of following
i) L = 7.2e
= 7.2 × 11
= 79.2 m ….(i)
ii) L = 0.073D × Vmax
= 0.073 × 7.6 × 90
= 49.93 m ….(ii)
here, D → cant deficiency for BG track is 7.6 cm
V2 V2
a) e = 1.676 b) e = 1.315
R R
V2 V2
c) e = 0.80 d) e = 0.60
R R
Sol: b) [2017, SET-II]
34
Chapter
Previous Years’ Question
For a 2° curve on a high speed Broad Gauge (BG) rail section, the
maximum sanctioned speed is 100 km/h, and the equilibrium speed
is 80 km/h. Consider the dynamic gauge of BG rail as 1750 mm. The
degree of curve is defined as the angle subtended at its centre by
a 30.5 m are. The cant deficiency for the curve (in mm, round off to
integer) is_____________.
Sol: 57 [2021-Set-II]
Keywords
35