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CE-363
Lecture – 5: Stresses in
Components of Track
Dr. Ankit Gupta, Assistant Professor
Department of Civil Engineering
National Institute of Technology Hamirpur
Lecture Outline
Stresses in Rails
Stresses in Sleepers
Stresses in Ballast
Stresses in Formation
Coning of Wheels
Tilting of Rails / Adzing of Sleepers
Stresses in Rails
The number of sleepers per rail have little effect on the stresses in rails.
Eccentric vertical loads causes bending moment along with torsion of beam.
Lateral thrust at the rail head produces lateral deflection and twisting of the rail
Lateral deflection also causes lateral movement of sleepers on either side.
Contact shear stress under repeated load condition causes metals of the wheel tread and rail gets fatigued
Stresses in Rails
Very high stresses occur in lower fillets on
sharp curves.
Temperature causes alteration of tensile
and compressive stresses.
Tractive effort and braking force results in
flow of metal and deformation of rail head.
The effect is higher with small diameter
wheels.
Track irregularities causes heavy impact
and deflection
Stresses in Rails
Bending stress is rails (due to vertical
loads)
Assumption is that rail is a long bar
continuously supported by an elastic
foundation.
On account of vertical loads the rail is
subjected to bending and flexural
stresses.
Based on the elastic theory, the bending
moment is computed as:
Stresses in Rails
Bending stress is rails (due to vertical
loads)
M = 0.25Ple-x/l {Sin (x/l) – Cos (x/l)}
where P = isolated vertical load
I = Characteristics length
= {4EI / µ}1\4
EI = Flexural stiffness of the rail
µ = Track Modulus
x = distance of the point under
consideration from the load
Stresses in Rails
Bending stress is rails (due to vertical
loads)
Bending Stress = BM / Sectional Modulus of rail
BM is zero at points x = πl/4, 3 πl/4 and is
maximum at x = 0, πl/2, 3 πl/2, etc.
+20
-20
-40
-60
-80
-
100
Πl/4 Πl/2 3Πl/4 Πl 5Πl/4
Length of track
Max BM = Pl/4
Stresses in Rails
Bending Moment can also be computed
using:
Mmax = 0.318 W . x (in cm .tons)
Where W = Isolated wheel load in tons
x = Distance from the load to the
point of contra flexure in cm
= 42.3 [I/µ]1/4 (in cm)
µ = sub-grade modulus in kg/cm2
I = Moment of inertia of rail section
in cm4
Stresses in Rails
Bending stress then can be computed as:
f = Mmax/Z (in kg/cm2) OR
f = 13.8 W/Z . [I/µ]1/4 (in tons/cm2)
Where Mmax = maximum BM in cm . tons
f = stress due to BM immediately
under the load W in kg/cm2
(permissible value 22 to 25 kg/mm2)
Z = Modulus of rail section in cm3
Stresses in Rails
Deflection due to vertical load can be
computed as:
d = 9.25 W/[Iµ3]1/4 (in cm)
Where d = deflection of track in cm
W = Isolated wheel load in tonne
µ = Modulus of track in kg/cm2
I = Moment of inertia of rail section
in cm4
Stresses in Rails
Reduction
Factors
Stresses in Sleepers
Various factors causing stresses in
sleepers are:
Dead and moving loads
Wheel load
Weight transfer from one wheel to another
wheel
Effect of speed on the rail and
Dynamic augment
Stresses in Sleepers
Factors causing stresses in sleepers
Track components
Track modulus
Elasticity of rail and its stiffness
Design and strength of sleeper
Sleeper density
Type and efficiency of fastenings
Stresses in Sleepers
Factors causing stresses in sleepers
Maintenance
Irregularities in rails
Their maintenance
Maintenance of wheels of coaches
Effectiveness of supervision for ensuring
good maintenance
Stresses in Sleepers
Distribution of stresses in sleepers
Due to alteration in ballast reaction under
sleepers the pattern of distribution of
stresses gets modified.
The wave motion of rail causes movement
of load on the rail seat from one edge of
the sleeper to the other and makes its
distribution non-uniform.
Stresses in Sleepers
Distribution of stresses in sleepers
End bound sleepers:
In case of newly constructed and
compacted sleepers, more deflection is
caused at the center as compared to the
ends. This is called end bound sleepers.
Load transferred
through rails
Sleepers
Stresses in Sleepers
Distribution of stresses in sleepers
Centre bound sleepers:
With the repeated application of loads a
depression is caused at the end of the
sleepers. This results in higher depression
at the ends as
compared to the centre.
Load transferred
through rails
Sleepers
Stresses in Sleepers
Distribution of stresses in sleepers
Load on Rail Seat:
Max. load on rail seat
= (P/Zµl) . µ . S
= P.S / Z . l
where P = wheel load
S = sleeper spacing
l = characteristics length
Z = Modulus of rail section
Stresses in Sleepers
Distribution of stresses in sleepers
This is approximately 30 to 50% of the dynamic wheel load.
The length of the sleeper and the rail seat is so selected that under normal loading, the middle part of the sleeper is stress less
Provision of bearing plates also reduces the maximum stress in sleepers by about 32% on BG track.
Stresses in Ballast
Depends up on Live load and dead load of superstructure
and trains
Elastic properties of the sleepers
Section and length of sleepers
Spacing of sleepers
Degree of compaction of ballast
Nature of ballast bed
Size, gradation, depth and compaction of ballast
Type of sub-grade
Pattern of distribution of load / stresses
Stresses in Ballast
It is observed that
Pressure on the sleeper is maximum at
the centre of its width and this pressure
decreases from centre towards the ends.
The vertical pressure under the sleeper is
uniform at a depth which is approximately
equal to the spacing of the sleepers.
Stresses in Ballast
Distribution of stresses / pressure in ballast
section
40 Percentage
of average
pressure
50
100
30
10
1
Depth of
ballast
Spacing of
sleeper
Line of uniform pressure
Stresses in Ballast
Distribution of stresses / pressure in ballast
section
Depth of
ballast (z)
Spacing of
sleeper
Bottom of
Tie
Pressure at the bottom of tie = Pa
Pressure at depth ‘z’ = Pz
Pressure at distance ‘x’ for depth ‘z’ =
Px
Pz
Px
Pa
P
Width of sleeper
(w)
Length of
sleeper
(L)
Stresses in Ballast
Distribution of stresses / pressure in ballast
section
Pressure at the bottom of the tie (sleeper) is
given by:
Pa = P/ (w . L/2) in kg/sq.m
Where, P = isolated wheel load, in kg
w = width of the tie or sleeper, in m
L = length of the sleeper, in m
Stresses in Ballast
Distribution of stresses / pressure in
ballast section
Pressure at depth ‘z’ (in cm) below centre of
tie width is computed as:
Pz = 5.24 Pa / z1.25 (in kg/sq.m)
Pressure at distance ‘x’ (in cm) at depth ‘z’ is
given by:
Px = 0.48 (Pa / z) . 10-2.06 (x / z) (x / z)
(in kg/sq.m)
Stresses in Sub-grade
Bearing pressure of sub-grade soil
Alluvial soil below 0.70 kg/cm2
Soft clay 1.12 – 1.41kg/cm2
Wet or loose sand 1.12 – 1.41 kg/cm2
Dry clay or firm sand 1.48 – 2.11 kg/cm2
Compacted soil 2.88 & over
Stresses in Sub-grade
Maximum formation pressure permitted
on Indian Railways
For motive power 3.5 kg/cm2 for BG
2.5 kg/cm2 for MG
For goods wagons 3.0 kg/cm2 for BG
2.3 kg/cm2 for MG
Permissible Values
Bending stress in rails 36.0 kg/mm2
Contact stress between
rail and wheel 21.6 kg/mm2
Dynamic over loads at rail joints due to unsuspended masses
27 tons for locos
19 tons for wagons
Formation pressure 3.5 kg/cm2 for locos
3.0 kg/cm2 for wagons
Permissible Values
Fish plate stresses 30 kg/mm2
Bolt hole stresses 27 kg/mm2
Minimum ultimate tensile strength
72 kg/mm2
Assumed yield point 42.5 kg/mm2
Coning of Wheels
Flat Surface Coned Surface
Coning of Wheels
Problems with flat wheel
Lateral Sway on straight track
Wearing of flanges and side of rail head
Unequal movement on curved rails
Longer distance to be moved on outer curved rail as compared to inner curved rail
Flexibility not available due to rigidity of vehicle base.
Coning of Wheels
Coning causes –
On a straight track
Bringing back wheel to average diameter by slipping the wheel
On a curved track
Shifting the outer wheel outwards (due to centrifugal force) thus causing an increase in diameter that helps it in moving longer distance on outer curve as compared to inner wheel for which the diameter reduces thus making it to traverse shorter distance.
Coning of Wheels
Coning of Wheels
Coning helps in –
Controlling differential movement of
front and rear axles caused due to
rigidity of frame and axle, thus acting
as a balancing factor.
On curves the rear axle has a tendency
to move towards inner rail.
Reducing wear and tear of wheel
flanges
Smooth riding.
Coning of Wheels
Problem with Coning of Wheels
Wear and tear due to slipping action
Slip of wheel = [2/360] x G
= angle made by rigid wheel base at
center of the curve
BG slip = 0.029 m per degree curve
Eccentric loading on rails
Coning of Wheels
Problem with Coning of Wheels
Coning of Wheels
RAIL
Tilting of Rails / Adzing of Sleepers
Rails are tilted at an angle of 1 in 20
Controls –
Lateral bending stresses due to
eccentric loading
Reduces wear and tear at inner edge of
rail and on tread of wheel
Adzing of sleepers or use of canted
bearing plates
Tilting of Rails / Adzing of Sleepers