RAILWAY ENGINEERING

&
RUNWAY ENGINEERING
Curve
• A curve is defined either by its radius or by
its degree.
• The degree of the curve is the angle
subtended at its centre by a 100 ft chord
or 30.5 m chord.
• The value of degree of curve can be
determined by
– Circumference of circle = 2  r
2
Curve (cont’d)
• Angle subtended at the centre by the circle
with this circumference = 360
• Angle subtended at the centre by a 30.5 m
chord or degree of curve
• D = 360* 30.5/ 2* pi*r

• D = 1750/R (R in meters)
• D = 5730/R ( R in feet)
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Problem
• Find out the radius of the track laid on 4 o
curve.

4
Problem

• Find out the degree of the curve if the

radius is 1500 ft.

5
Problem

• Find out the degree of the curve if the

radius is 500 m.
•

6
Problem

• Find out the radius of track in ft laid on 3

degree of the curve.

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Why Curves are provided
• It is desirable to lay the track as straight
and gentle as possible but it is not
possible due to
– Natural features of the country
– Due to necessity of avoiding obstruction both
natural and artificial
• As curves are unavoidable so it is
desirable to lay track on the curves as flat
as possible.
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Why Curves should be avoided
• Curves produce resistance to haulage of
trains
• Wear both in track and vehicle
• Reduce the safe speed limits
• Increase the maintenance cost

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 Compensation for Curvature
• Extra power is required to move the train
along the curve.
• If speed of the train is not to be reduced
that is the effective pulling power is to be
maintained.
• Extra power required is made available by
the reducing the extra effort required to
negotiate the gradient on the curve.

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 Compensation for Curvature
• When sharp curve is to laid, gradient is kept
smaller than the ruling gradient and the amount
by which it is reduced is compensation of
curvature
• When sharp curves are provided there is speed
reduction
– Due to Curvature
– Also extra power is required to overcome increased
frictional resistance due to centrifugal force and
sliding.
• Normally compensation of 0.03 % to 0.05 % is
provided for every degree of curvature in curve.
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Example
• If a 3 degree curve is located on a ruling
gradient of 1 in 150. How much is the
gradient on the curve
• Sol
Gradient = 1/150*100 =0.67%
Assuming grade compensation = 0.05% for 1
degree.
Grade compensation for 3 deg = 0.05*3=0.15
0.67 – (3* 0.05) = 0.52% or 1 in 192.
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Problem
• Find out the actual gradient of the track
laid on 4 o curve. The track is Broad gauge
track and ruling gradient is 1 in 200.

13
Problem
• Find out the actual gradient of the track
laid on 5 o curve. The track is Standard
gauge track and ruling gradient is 1 in 150.

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Types of Curves
• Circular Curves
• Transition
• Vertical Curves

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Circular Curve
• Curve of constant radius and can be
described by the radius or degree of
curvature.
• Circular curves are classified as
– Simple curve- curve of constant radius
– Compound curve- circular curves of two
different radii but curving in the same direction
– Reverse curve- formed by two circular curves
each curving in the opposite direction

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Curves Types (cont’d)
• Transition Curves
– Curve of variable radius and provided
between straight and circular curve
– Also provided between two circular curves of
different radii
– They are also called easement curves
– Cubic parabola is used for transition curves
• Vertical Curves
– Provided at intersection of two gradient lines

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Variable affecting Speed on Curves
• When the train moves over the curve
centrifugal force is exerted on the outer rail
• Magnitude of the force depends upon
speed of train and radius of curve
• Other factors are
– Super elevation of track
– Rate of change of super elevation
– Super elevation gradient
– Super elevation deficiency and excess cant
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Super elevation or Cant
G

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BC - is the super elevation
W- axle load
P- centrifugal force
G- gauge of the track
F- Resultant of the two forces to pass through centre E
of the track to load the two tracks equally

W sin   P cos 
P
tan  
W 20
Let v = speed of train in miles per hour/ km per hour
R = radius of the curve in ft or in meters
g = acceleration due to gravity
P = centrifugal force acting on the vehicle through its centre of
gravity

Centrifugal Force is given by

2
WV
P
gR 21
 Substituting the value of P in Eq. 1

WV 2 1
tan    (2)
gR W

BC
tan   (3)
AB

Substituting the value of tan  in Eq. 2

V2
BC   ( AB)
gR

AB  G (gauge of track)

BC= superelevation

GV 2
e
gR
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 Super elevations in terms of R
and V
• Calculate the expressions for BG, MG,
Standard gauge and narrow gauge track
when speed is in miles per hour and R is
in ft and super elevation is in inches

• 1 mile = 5280 ft
• g= 32.2 ft/sec2

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Expression for BG, SG and NG
• e = 4.4 V2/ R for BG

• e = 2.62 V2/ R for MG

• e = 2.0 V2/ R for NG

• R is in ft and speed in miles / hr

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 Super elevations in terms of R
and V
• Calculate the expressions for BG, MG,
Standard gauge and narrow gauge track
when speed is in km per hour and R is in
m and super elevation is in cm.

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 Problem
• Find SE of a 3o curve for a BG, MG, NG
track. The permitted speed is 50 mph.

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 Problem
• Find SE of a 4o curve for a BG, MG, NG
track. The permitted speed is 60 km/hr.

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Definition of Various Terms

• Cant/ Super elevation

– Amount by which one gauge rail is raised above the
level of the other for neutralizing the effect of
centrifugal force
– It is also called super – elevation
– It is positive when outer rail is above the inner rail and
negative when inner rail is above the outer rail
• Equilibrium Speed
– is the speed at which the vehicle is not subjected to
any un balanced centrifugal force.
– On a curve having a definite super-elevation

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Definition of Various Terms

• Super elevation Gradient

– They indicate increase or decrease in super
elevation.
• Rate of change of super elevation
– The rate at which super elevation is
decreased or increased relative to the
maximum speed of the vehicle passing over
transition curve.

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Definition of Various Terms

• Super elevation Deficiency

– When cant is provided for low speed and the train is
moving at a high speed, the existing cant is not
sufficient
– Amount by which existing cant is less than the
required to attain equilibrium that is cant deficiency
– Outer rail will be stressed
• Super elevation Excess
– When the super elevation is provided for high speed
and there is excess for slow moving train
– Heavy wear for inner rails

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Disadvantages of Avoiding Super
elevation

• Rapid wear of rails

• Forces the track out of alignment
• Possibilities of derailment

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 Negative Super elevation
• When a branch line on the curve joins the
main line
– It is not possible to provide super elevation on
the main line required for the design speed.
– Thus negative super elevation is provided on
the branch line
– Reduction in super elevation on the main line

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Negative Super Elevation
Main Line is ABCD
AB outer rail
CD inner rail

Branch Line AEFC

CF- outer rail
AE- inner rail
Point P should be higher than A for 33
Branch track but it is lower than A
Deficiency in Super Elevation and
Negative Super Elevation
• When diverging track, curves in the direction
opposite to that of the main curve, the super-
elevation necessary for the average speed of
the trains running over the main curve cannot be
given.

• The speed of trains over the diverging track has

to be considerably reduced, and the speed on
the main track may also have to be reduced.

• The reason for these reductions is that, on the

diverging track, the inner rail becomes higher
than the outer rail. 34
 Maximum SE permitted
• For Pakistan Deficiency in SE
– BG – 6.5 inches
BG- 3 inches
– MG- 4 inches
MG- 2 inches
– NG- 3 inches
NG – 2 inches

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 Problem
• A 6o curve branches off a 3o main curve in
opposite direction in the layout of a BG
track. If the speed on the branch track is
restricted to 22 mph. Determine speed
restriction on main line. Assume deficiency
in super elevation is 3 inches.

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Formation
• Formation is a level surface on which the ballast is
laid.
• The formation has to take all the loads coming on
the track.
• The top surface of the formation is kept slightly
sloping towards the sides, inorder to facilitate
drainage.
• Level of the formation is the level prepared at its
centre.
• Formation may be in fill or in cut or partially in fill
and partially in cut. When the section is in cut the
drainage becomes must.

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Function of Formation

• It distributes the load over wider area of

the natural ground.
• Provides a smooth level surface for the
placing of the ballast
• It facilitates the drainage
• It acts as foundation to the track and
provides stability to it.

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 Width of Formation
• Top width of the formation will depend upon
• Gauge of the track
• No of tracks
• Centre to centre distance between the tracks
• Width of the ballast layer
• Width of the drain on either side if the track is in
cutting
• Width of the formation for single lane BG track is
20 ft
• Width of the formation for double lane BG track
is 35 ft.
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40
Height of Formation

• Height of the formation depends upon the

– Soil conditions
– Highest flood level
– It also depends upon the natural topography/
topography.
– Intensity of pressure transmitted to the
ground.

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Side Slopes

• Side slope of the formation must be flatter

than the angle of repose of the ground.
• Side slopes in case of embankment is 1 to
2
• Side slopes in case of cutting is 1 to 1.5 or
even steeper
• The side is vertical when solid rock exists.
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Formation Width for Embankment
• Gauge (Broad Gauge)
– Single 20 ft (6.10 m)
– Double 35.5 ft (10.8 m)
• Meter Gauge
– Single 16 ft (4.9 m)
– Double 29 ft (8.8 m)
• Narrow Gauge
– Single 12 ft (3.7 m)
– Double 24 ft (7.3 m)

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Formation Width for Cutting
• Gauge (Broad Gauge)
– Single 18 ft(5.5 m)
– Double 35.5 ft (10.2 m)
• Meter Gauge
– Single 14 ft (4.3 m)
– Double 27 ft (8.3 m)
• Narrow Gauge
– Single 11 ft (3.4 m)
– Double 23 ft (7.0 m)

– Note c/c distance for track is 15 ft for BG

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