railway engineering

1. DEBRE MARKOS UNIVERSITY
INSTITUTE OF TECHNOLOGY
SCHOOL OF CIVIL AND WATER RESOURCE ENGINEERING
Railway Engineering
Biniyam A 2022

2. Track gauge
Track Gauge:
• Definition: Track gauge, G, is the smallest distance between lines
perpendicular to the running surface intersecting each rail head profile at
point P in a range from 0 to Z_p below the running surface.
Z_p is always 14 mm
In the situation of new unworn rail head the point P will be at the limit Z_p
below the railhead

3. Track gauge
• Distance between the inner sides of the rails, measured 14 mm
below rolling surface

4. Track gauge
Different track gauge values (e)
• Standard (Normal) gauge (e = 1435 mm); Major parts of European
railways, USA, Canada, Mexico, China use standard gauge.
• Metric (Narrow) gauge (e = 1000 mm or e = 1067mm);
• Broad gauge (e = 1524 mm or 1672 mm); Finland, Spain, Russia and
countries in the Former Soviet Union, India, etc

5. Track gauge
Track gauges around the world

6. Track gauge
• All track gauges give the same operational possibilities, however the
metric gauges don’t give high speed operation
• Track gauge and loading gauge are parameters that determine rail network
compatibility with adjacent networks, present or future, but the economy
aspect of track gauge is minimum

7. Track Gauge
• EN 13848-1:2003+A1:2008: Track Gauge
In the situation of used worn rail head the point P for the left rail can be
different from the right rail.

8. Track gauge
The theoretical track gauge can be
varied due to:
• Rail wear
• Rail head deformation
• Rail roll (rail tilt)
• Loose fastening
• Rail buckling
• EN standards allow for some deviation of the
track gauge from the theoretical value – track
gauge tolerances
• Track gauge tolerances are limited in turnouts

9. Track gauge widening
National standards may require gauge widening in small radius
curves
EN (UIC) standard:
• 10 mm for curve radius 150-199 m
• 15 mm for curve radius 120-149 m
• 20 mm for curve radius < 120 m
Gauge widening application depends on the type of sleeper and the
track form
Gauge narrowing (tightening) has very small
tolerance -5 mm (gauge value 1430 mm) …
safety risk

10. Track gauge widening
The designed gauge should change linearly
• Full gauge widening should be provided for the whole circular curve with
the small radius
• The gauge transition (1 mm per meter) must be placed on the adjacent
horizontal element (straight track, circular curve with larger radius, or –
ideally – a transition curve)
• Gauge widening should be avoided in turnouts. Hence, no turnouts within
10-15 metres from a small radius curve

11. Track gauge
Measurement technique
• Manual measurement method unloaded track
• Using Track recording car - loaded track (Roger
1000 – 15 tonnes axle load)
Consequences of track gauge variations:
• Gauge too wide  derailment
• Gauge too narrow  flange climbing (derailment)
• Varying gauge  poor lateral stability
• Gauge too wide and reverse curves  buffer
locking

12. Wheel-rail contact mechanics

13. Wheelset definitions

14. • RegionA: wheel tread-rail head
contact
– most common contact region
– lower contact stress
• Region B: wheel flange-rail gauge
corner
– much smaller contact area and more
severe
– higher contact stresses and wear rates
• Region C: wheel and rail field sides
contact
– Least likely contact region
– High contact stress
– Undesirable wear lead to incorrect
steering of wheelset
Wheel-rail contact regions

15. Phenomena in the wheel-rail interface
• High vertical, lateral and
longitudinal contact forces,
induce stresses that may
cause material yielding and
fatigue
• Rolling contact forces
combined with friction
induce wear
• Traction and braking lead to
wheel sliding leading to
- rail burns and wheel flats
- thermal cracks

16. Phenomena in the wheel-rail interface
• These phenomena may cause
irregularities
– poor vehicle dynamics
– further increase in contact forces
– increase in vibration and noise
• Consequences
– discomfort for passengers
– disturbance for the surrounding
– increased maintenance cost for wheel and
rails and other components
– in a sever case to derailment induced by
wheel and rail fracture or by the wheel
flange climbing on the rail

17. Basics Concepts
The wheels are made conical, the smaller circumference at the outer edge.
…if anything throws the wheels in the slightest degree to one side the wheel
is immediately rolling on a larger circumference than the other and the
tendency to roll back is introduced. The carriage is kept always in the
middle of the track. A beautiful arrangement.
Brunel, 1838
The flanges are a necessary precaution but they ought never to touch the
rail and therefore they cannot be said to keep the wheels on the rails. They
ought not to come into action except to meet an accidental, lateral force.….

18. Definitions
The steering mechanism of the
wheelset is due to the equivalent
conicity (the rolling radius difference
between the left and the right wheels)

19. Motion of a free wheelset
y
2b
ro-λ.y
ro+λ.y
An idealised conical wheelset displaced laterally on cylindrical rails
rl = ro-λ.y rr = ro+λ.y

20. Perfect Curving
For perfect curving:
r0 y

R b
r0 y R b
Where r0 = the radius when the wheelset is central
b = half the gauge
R = the radius of the curve
 = the conicity of wheel tread (inclination)
so:
R
y 
r0b
• If the flangeway clearance is exceeded then
perfect curving cannot occur and flange
contact will take place

21. Example
1.Awheel with straight cones will have a conicity the same as the cone
inclination tan𝜆 = 𝜆.
For the following data find the lateral displacement needed to achieve a
perfect radial steering: rolling radius r = 0.5, 𝜆=0.05 (inclination 1:20), b =
0.75 and R = 500m
• For a perfect curving
R
r0  y R  b
r0  y

R  b y 
r0b

 0.015m  15mm
0.05 500
y 
r0b

R
0.5 0.75
y 
15 mm lateral displacement is
needed to achieve perfect radial
steering

22. Example
2. The cone inclination is instead 𝜆=0.25 (inclination 1:4), find the lateral
displacement needed to achieve a perfect radial steering
• For a perfect curving
y 
r0b
R
y 
0.5

0.75
 0.003m  3mm
0.25 500
Only 3 mm lateral displacement would be
sufficient to achieve perfect radial steering
r  2  0.25 0.003
r  1.5 mm
 r  2y
r
2y
 
r  rl  rr
A rolling radius difference
between the outer and the inner
wheel of 1.5 mm needed